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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Show that the points $(\hat{i} - \hat{j} + \hat{k})$ and 3 $(\hat{i} + \hat{j} + \hat{k})$ are equidistant from the plane and lie on opposite sides of it.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e P (\hat{i} - \hat{j} + 3 \hat{k}) a n d Q (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) a n d t h e p l a n e \vec{r} . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9 = 0 \\ P e r p e n d i c u l a r distance o f P (\hat{i} - \hat{j} + 3 \hat{k}) f r o m t h e p l a n e \\ \vec{r} . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9 = | \frac{(\hat{i} - \hat{j} + 3 \hat{k}) . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9}{\sqrt[]{{(5)}^{2} + {(2)}^{2} + {(- 7)}^{2}}} | \\ = | \frac{5 - 2 - 21 + 9}{\sqrt[]{25 + 4 + 49}} | = | \frac{- 9}{\sqrt[]{78}} | \\ a n d p e r p e n d i c u l a r distance o f Q (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) f r o m t h e p l a n e \\ = | \frac{(3 \hat{i} + 3 \hat{j} + 3 \hat{k}) . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9}{\sqrt[]{25 + 4 + 49}} | \\ = | \frac{15 + 6 - 21 + 9}{\sqrt[]{78}} | = | \frac{9}{\sqrt[]{78}} | \\ H e n c e, t h e t w o points a r e equidistant f r o m t h e g i v e n p l a n e . \\ O p p o s i t e s i g n s h o w s t h a t t h e y l i e o n e i t h e r s i d e o f t h e p l a n e . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the equation of the plane through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3 \hat{j}) - 6 = 0 and \vec{r} \cdot (3 \hat{i} - \hat{j} - 4 \hat{k}) = 0,$ whose perpendicular distance from the origin is unity.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n p l a n e s a r e : \\ \vec{r} . (\hat{i} - 3 \hat{j}) - 6 = 0 \Rightarrow x + 3 y - 6 = 0 \dots (i) \\ a n d \vec{r} . (3 \hat{i} - \hat{j} - 4 \hat{k}) = 0 \Rightarrow 3 x - y - 4 z = 0 \dots (i i) \\ E q u a t i o n o f t h e p l a n e passing t h r o u g h t h e l i n e o f intersection o f p l a n e (i) a n d (i i) i s \\ (x + 3 y - 6) + k (3 x - y - 4 z) = 0 \dots (i i i) \\ (1 + 3 k) x + (3 - k) y - 4 k z - 6 = 0 \\ P e r p e n d i c u l a r distance f r o m o r i g i n \\ \Rightarrow | \frac{- 6}{\sqrt[]{{(1 + 3 k)}^{2} + {(3 - k)}^{2} + {(- 4 k)}^{2}}} | = 1 \\ \Rightarrow \frac{36}{1 + 9 k^{2} + 6 k + 9 + k^{2} - 6 k + 16 k^{2}} = 1 [S q u a r i n g b o t h s i d e s] \\ \Rightarrow \frac{36}{26 k^{2} + 10} = 1 \Rightarrow 26 k^{2} + 10 = 36 \\ \Rightarrow 26 k^{2} = 26 \Rightarrow k^{2} = 1 ∴ k = \pm 1 \\ P u t t i n g t h e v a l u e o f k i n e q . (i i i) w e g e t, \\ (x + 3 y - 6) \pm (3 x - y - 4 z) = 0 \\ \Rightarrow x + 3 y - 6 + 3 x - y - 4 z = 0 a n d x + 3 y - 6 - 3 x + y + 4 z = 0 \\ \Rightarrow 4 x + 2 y - 4 z - 6 = 0 a n d - 2 x + 4 y + 4 z - 6 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n a r e \\ 4 x + 2 y - 4 z - 6 = 0 a n d - 2 x + 4 y + 4 z - 6 = 0 \end{array}$

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3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

The plane $a x + b y = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $α$ . Prove that the equation of the plane in its new position is $a x + b y \pm \sqrt[]{a^{2} + b^{2}} t a n (α) z = 0 .$

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n p l a n e s a r e \\ a x + b y = 0 \dots (i) \\ z = 0 \dots (i i) \\ E q u a t i o n o f a n y p l a n e passing t h r o u g h t h e l i n e o f intersection o f p l a n e (i) a n d (i i) i s \\ (a x + b y) + k z = 0 \Rightarrow a x + b y + k z = 0 \dots (i i i) \\ D i v i d i n g b o t h s i d e s b y \sqrt[]{a^{2} + b^{2} + k^{2}}, w e g e t \\ \frac{a}{\sqrt[]{a^{2} + b^{2} + k^{2}}} x + \frac{b}{\sqrt[]{a^{2} + b^{2} + k^{2}}} y + \frac{k}{\sqrt[]{a^{2} + b^{2} + k^{2}}} z = 0 \\ ∴ D i r e c t i o n cosines o f t h e n o r m a l t o t h e p l a n e a r e \\ \frac{a}{\sqrt[]{a^{2} + b^{2} + k^{2}}}, \frac{b}{\sqrt[]{a^{2} + b^{2} + k^{2}}}, \frac{k}{\sqrt[]{a^{2} + b^{2} + k^{2}}} \\ a n d t h e d i r e c t i o n cosines o f t h e p l a n e (i) a r e \\ \frac{a}{\sqrt[]{a^{2} + b^{2}}}, \frac{b}{\sqrt[]{a^{2} + b^{2}}}, 0 \\ Since, α i s t h e a n g l e b e t w e e n t h e p l a n e s (i) a n d (i i i), w e g e t \\ c o s α = \frac{a . a + b . b + k . 0}{\sqrt[]{a^{2} + b^{2} + k^{2}} . \sqrt[]{a^{2} + b^{2}}} \\ \Rightarrow c o s α = \frac{a^{2} + b^{2}}{\sqrt[]{a^{2} + b^{2} + k^{2}} . \sqrt[]{a^{2} + b^{2}}} \\ \Rightarrow c o s α = \frac{\sqrt[]{a^{2} + b^{2}}}{\sqrt[]{a^{2} + b^{2} + k^{2}}} \Rightarrow c o s^{2} α = \frac{a^{2} + b^{2}}{a^{2} + b^{2} + k^{2}} \\ \Rightarrow (a^{2} + b^{2} + k^{2}) c o s^{2} α = a^{2} + b^{2} \\ \Rightarrow a^{2} c o s^{2} α + b^{2} c o s^{2} α + k^{2} c o s^{2} α = a^{2} + b^{2} \\ \Rightarrow k^{2} c o s^{2} α = a^{2} - a^{2} c o s^{2} α + b^{2} - b^{2} c o s^{2} α \\ \Rightarrow k^{2} c o s^{2} α = a^{2} (1 - c o s^{2} α) + b^{2} (1 - c o s^{2} α) \\ \Rightarrow k^{2} c o s^{2} α = a^{2} s i n^{2} α + b^{2} s i n^{2} α \\ \Rightarrow k^{2} c o s^{2} α = (a^{2} + b^{2}) s i n^{2} α \\ \Rightarrow k^{2} = (a^{2} + b^{2}) \frac{s i n^{2} α}{c o s^{2} α} \Rightarrow k = \pm \sqrt[]{a^{2} + b^{2}} . t a n α \\ P u t t i n g t h e v a l u e o f k i n e q . (i i i) w e g e t \\ a x + b y \pm (\sqrt[]{a^{2} + b^{2}} . t a n α) z = 0 w h i c h i s t h e r e q u i r e d E q u a t i o n o f p l a n e . \\ H e n c e p r o v e d . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the equation of the plane which is perpendicular to the plane $5 x + 3 y + 6 z + 8 = 0$ and contains the line of intersection of the planes $x + 2 y + 3 z – 4 = 0$ and $2 x + y – z + 5 = 0$ .

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n p l a n e s a r e \\ P_{1} : 5 x + 3 y + 6 z + 8 = 0 \\ P_{2} : x + 2 y + 3 z - 4 = 0 \\ P_{3} : 2 x + y - z + 5 = 0 \\ E q u a t i o n s o f t h e p l a n e passing t h r o u g h t h e l i n e o f intersection o f P_{2} a n d P_{3} i s \\ (x + 2 y + 3 z - 4) + λ (2 x + y - z + 5) = 0 \\ \Rightarrow (1 + 2 λ) x + (2 + λ) y + (3 - λ) z - 4 + 5 λ = 0 \dots (i) \\ P l a n e (i) i s p e r p e n d i c u l a r t o P_{1}, t h e n \\ 5 (1 + 2 λ) + 3 (2 + λ) + 6 (3 - λ) = 0 \\ \Rightarrow 5 + 10 λ + 6 + 3 λ + 18 - 6 λ = 0 \\ \Rightarrow 7 λ + 29 = 0 \Rightarrow λ = \frac{- 29}{7} \\ P u t t i n g t h e v a l u e o f λ i n e q . (i), w e g e t \\ [1 + 2 (\frac{- 29}{7})] x + [2 - \frac{29}{7}] y + [3 + \frac{29}{7}] z - 4 + 5 (\frac{- 29}{7}) = 0 \\ \Rightarrow \frac{- 15}{7} x - \frac{15}{7} y + \frac{50}{7} z - 4 - \frac{145}{7} = 0 \\ \Rightarrow - 15 x - 15 y + 50 z - 28 - 145 = 0 \\ \Rightarrow - 15 x - 15 y + 50 z - 173 = 0 \Rightarrow 15 x + 15 y - 50 z + 173 = 0 \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the shortest distance between the lines $\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}$ and $\vec{r} = 15 \hat{i} - 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})$

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n s o f l i n e s a r e \\ \vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k} \dots (i) \\ a n d \vec{r} = 15 \hat{i} - 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k}) \dots (i i) \\ E q u a t i o n (i) c a n b e r e - w r i t t e n a s \\ \vec{r} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k} + λ (3 \hat{i} - 16 \hat{j} + 7 \hat{k}) \dots (i i i) \\ H e r e, \vec{a_{1}} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k} a n d \vec{a_{2}} = 15 \hat{i} - 29 \hat{j} + 5 \hat{k} \\ \vec{b_{1}} = 3 \hat{i} - 16 \hat{j} + 7 \hat{k} a n d \vec{b_{2}} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \\ \vec{a_{2}} - \vec{a_{1}} = 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \\ \vec{b_{1}} * \vec{b_{2}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 16 & 7 \\ 3 & 8 & - 5 \end{matrix} | \\ = \hat{i} (80 - 56) - \hat{j} (- 15 - 21) + \hat{k} (24 + 48) \\ = 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \\ ∴ S h o r t e s t distance, S D = | \frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} * \vec{b_{2}})}{| \vec{b_{1}} * \vec{b_{2}} |} | = | \frac{(7 \hat{i} + 38 \hat{j} - 5 \hat{k}) . (24 \hat{i} + 36 \hat{j} + 72 \hat{k})}{\sqrt[]{{(24)}^{2} + {(36)}^{2} + {(72)}^{2}}} | \\ = | \frac{168 + 1368 - 360}{\sqrt[]{576 + 1296 + 5184}} | = | \frac{168 + 1008}{\sqrt[]{7056}} | = \frac{1176}{84} = 14 u n i t s \\ H e n c e, t h e r e q u i r e d distance i s 14 u n i t s . \end{array}$

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3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane $x - 2 y + 4 z = 10$ .

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} E q u a t i o n o f p l a n e passing t h r o u g h t w o points (x_{1}, y_{1,} z_{1}) a n d (x_{2}, y_{2}, z_{2}) w i t h i t s n o r m a l d' r a t i o s i s \\ a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0 \dots (i) \\ I f t h e p l a n e i s passing t h r o u g h t h e g i v e n points (2, 1, - 1) a n d (- 1, 3, 4) t h e n \\ a (x_{2} - x_{1}) + b (y_{2} - y_{1}) + c (z_{2} - z_{1}) = 0 \\ \Rightarrow a (- 1 - 2) + b (3 - 1) + c (4 + 1) = 0 \\ \Rightarrow - 3 a + 2 b + 5 c = 0 \dots (i i) \\ Since, t h e r e q u i r e d p l a n e i s p e r p e n d i c u l a r t o t h e g i v e n p l a n e x - 2 y + 4 z = 10, t h e n \\ 1 . a - 2 . b + 4 . c = 10 \dots (i i i) \\ S o l v i n g (i i) a n d (i i i) w e g e t, \\ \frac{a}{8 + 10} = \frac{- b}{- 12 - 5} = \frac{c}{6 - 2} = λ \\ a = 18 λ, b = 17 λ, c = 4 λ \\ H e n c e, t h e r e q u i r e d p l a n e i s 18 λ (x - 2) + 17 λ (y - 1) + 4 λ (z + 1) = 0 \\ \Rightarrow 18 x - 36 + 17 y - 17 + 4 z + 4 = 0 \\ \Rightarrow 18 x + 17 y + 4 z - 49 = 0 \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes $x + 2 y = 0$ and $3 y - z = 0$ .

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n point i s (3, 0, 1) a n d t h e e q u a t i o n o f p l a n e s a r e \\ x + 2 y = 0 \dots (i) \\ a n d 3 y - z = 0 \dots (i i) \\ E q u a t i o n o f a n y l i n e l passing t h r o u g h (3, 0, 1) i s \\ l : \frac{x - 3}{a} = \frac{y - 0}{b} = \frac{z - 1}{c} \\ D i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e (i) a n d (i i) a r e (1, 2, 0) a n d (0, 3, - 1) \\ Since, t h e l i n e i s p a r a l l e l t o b o t h t h e p l a n e s . \\ ∴ 1 . a + 2 . b + 0 . c = 0 \Rightarrow a + 2 b + 0 c = 0 \\ a n d 0 . a + 3 . b - 1 . c = 0 \Rightarrow 0 . a + 3 b - c = 0 \\ S o, \frac{a}{- 2 - 0} = \frac{- b}{- 1 - 0} = \frac{c}{3 - 0} = λ \\ ∴ a = - 2 λ, b = λ, c = 3 λ \\ S o, e q u a t i o n o f l i n e i s \\ \frac{x - 3}{- 2 λ} = \frac{y - 0}{λ} = \frac{z - 1}{3 λ} \\ H e n c e, t h e r e q u i r e d e q u a t i o n \frac{x - 3}{- 2} + \frac{y - 0}{1} + \frac{z - 1}{3} \\ o r i n v e c t o r f o r m i s (x - 3) \hat{i} + y \hat{j} + (z - 1) \hat{k} = λ (- 2 \hat{i} + \hat{j} + 3 \hat{k}) \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the length and the foot of the perpendicular from the point ( $1, \frac{3}{2}, 2$ ) to the plane $2 x - 2 y + 4 z + 5 = 0$ .

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n p l a n e i s 2 x - 2 y + 4 z + 5 = 0 a n d t h e g i v e n points i s (1, \frac{3}{2}, 2) \\ D' r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e 2, - 2, 4 \\ S o, t h e ? e q u a t i o n o f l i n e passing t h r o u g h (1, \frac{3}{2}, 2) a n d w h o s e d' r a t i o s a r e e q u a l t o t h e \\ D' r a t i o s o f t h e n o r m a l t o t h e p l a n e i . e ., 2, - 2, 4 i s \frac{x - 1}{2} = \frac{y - \frac{3}{2}}{- 2} = \frac{z - 2}{4} = λ \\ ∴ A n y point i n t h e p l a n e i s 2 λ + 1, - 2 λ + \frac{3}{2}, 4 λ + 2 \\ Since, t h e point l i e s i n t h e p l a n e, t h e n \\ 2 (2 λ + 1) - 2 (- 2 λ + \frac{3}{2}) + 4 (4 λ + 2) + 5 = 0 \\ \Rightarrow 4 λ + 2 + 4 λ - 3 + 16 λ + 8 + 5 = 0 \\ \Rightarrow 24 λ + 12 = 0 ∴ λ = - \frac{1}{2} \\ S o, t h e c o o r d i n a t e s o f t h e point i n t h e p l a n e a r e \\ 2 (- \frac{1}{2}) + 1, - 2 (- \frac{1}{2}) + \frac{3}{2}, 4 (- \frac{1}{2}) + 2 i . e ., 0, \frac{5}{2}, 0 \\ H e n c e, t h e f o o t o f t h e p e r p e n d i c u l a r i s (0, \frac{5}{2}, 0) a n d t h e \\ r e q u i r e d l e n g t h = \sqrt[]{{(1 - 0)}^{2} + {(\frac{3}{2} - \frac{5}{2})}^{2} + {(2 - 0)}^{2}} \\ = \sqrt[]{1 + 1 + 4} = \sqrt[]{6} u n i t s \end{array}$

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3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the distance of the point (2, 4, –1) from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{– 9} .$

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n e q u a t i o n o f l i n e i s \\ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{– 9} = λ a n d a n y point P (2, 4, - 1) \\ L e t Q b e a n y point o n t h e g i v e n l i n e \\ ∴ C o o r d i n a t e s o f Q a r e x = λ - 5, y = 4 λ - 3 a n d z = - 9 λ + 6 \\ a n d t h e g i v e n point i s P (2, 3, - 8) \\ D i r e c t i o n r a t i o s o f P Q a r e λ - 5 - 2, 4 λ - 3 - 4, - 9 λ + 6 + 1 \\ i . e ., λ - 7, 4 λ - 7, - 9 λ + 7 \\ a n d t h e D' r a t i o s o f t h e g i v e n l i n e a r e 1, 4, - 9 . \\ I f P Q ⊥ l i n e \\ t h e n 1 (λ - 7) + 4 (4 λ - 7) - 9 (- 9 λ + 7) = 0 \\ \Rightarrow λ - 7 + 16 λ - 28 + 81 λ - 63 = 0 \\ \Rightarrow 98 λ - 98 = 0 \Rightarrow λ = 1 \\ S o, C o o r d i n a t e s o f Q a r e 1 - 5, 4 * 1 - 3, - 9 * 1 + 6 \\ i . e ., - 4, 1, - 3 \\ N o w, distance P Q = \sqrt[]{{(- 4 - 2)}^{2} + {(1 - 4)}^{2} + {(- 3 + 1)}^{2}} \\ = \sqrt[]{{(- 6)}^{2} + {(- 3)}^{2} + {(- 2)}^{2}} = \sqrt[]{36 + 9 + 4} = \sqrt[]{49} = 7 \\ H e n c e, t h e r e q u i r e d distance 7 u n i t s . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the distance of the point (2, 4, –1) from the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$

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