Three Dimensional Geometry

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4

Line are coplanar
so | [α, 5-α, 1], [2, -1, 1], | = 0
−5α + (α – 5)3 + 7 = 0
-2α = 8 ⇒ α = −4
⇒ L? : (x+2)/-4 = (y+1)/9 = (z+1)/1
Now by cross checking option (A) is correct.

w = I + j - 3k
p = 3i - j + k
q = -3i + 2j + 4k
p * q = |i, j, k; 3, -1, 1; -3, 2, 4| = -6i - 15j + 3k
S.D. = |AD. (p * q)|/|p * q| = | (36+225+9)|/√ (36+225+9) = 3√30

Determinant of vectors must be zero. Vector between points on lines: (-1-k, -2-2, -3-3). Vector directions: (1,2,3) and (3,2,1).
| -1-k, -4, -6; 1, 2, 3; 3, 2, 1 | = 0.
(-1-k) (2-6) - (-4) (1-9) + (-6) (2-6) = 0.
4 (1+k) - 32 + 24 = 0.
4+4k - 8 = 0. 4k=4 ⇒ k=1.

Vector on plane: (3-2, 7-3, -7- (-2) = (1,4, -5).
Line direction vector (-3,2,1).
Normal to plane n = (1,4, -5)* (-3,2,1) = (14,14,14) or (1,1,1).
Plane: 1 (x-3)+1 (y-7)+1 (z+7)=0 ⇒ x+y+z-3=0.
d = |-3|/√3 = √3. d²=3.

Normal to the required plane is perpendicular to the normals of the given planes.
n = n?  * n?  = (2i + j - k) * (i - j - k) = -2i + j - 3k.
Equation of the plane is -2 (x+1) + 1 (y-0) - 3 (z+2) = 0
-2x - 2 + y - 3z - 6 = 0
-2x + y - 3z - 8 = 0
2x - y + 3z + 8 = 0
Comparing with ax + by + cz + 8 = 0, we get a=2, b=-1, c=3.
a+b+c = 2-1+3 = 4.

ai+aj+ck,  i+k and ci+cj+bk are co-planar,
|a c; 1 0 1; c b| = 0
a (0-c) - a (b-c) + c (c-0) = 0
-ac - ab + AC + c² = 0
c² = ab
c = √ab

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

=> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21*\frac{34}{7}=102$

Now equation of line OA be

$\frac{x-1}{4}=\frac{y-3}{-5}=\frac{z-5}{2}=\lambda$

direction cosines of plane are 4, -5, 2

Equation of any point on OA be

$O\left(4\lambda +1,-5\lambda +3,2\lambda +5\right)$

Since O lies on given plane so

$4\left(4\lambda +1\right)-5\left(-5\lambda +3\right)+2\left(2\lambda +5\right)=8$

So, O (9/5,2,27/5). Hence by mid-point formula

B $\left(\frac{13}{5},1,\frac{29}{5}\right)\Rightarrow 5\left(\alpha +\beta +\gamma \right)=47$

Let $A\left(-2,-21,29\right),B\left(-1,-16,23\right),P\left(\lambda ,2,1\right),Q\left(4,-2,2\right)$

Given $\overrightarrow{AB}\perp \overrightarrow{PQ}$

$\therefore \overrightarrow{AB}.\overrightarrow{PQ}=0$

$\left(\widehat{i}+5\widehat{j}-6\widehat{k}\right).\left(\left(4-\lambda \right)\widehat{i}-4\widehat{j}+\widehat{k}\right)=0$

$4-\lambda -20-6=0$

$\lambda =-22$

$\therefore {\left(\frac{\lambda }{11}\right)}^2+\left(\frac{-4\lambda }{11}\right)-4=4+8-4=8$