Three Dimensional Geometry

$\Delta =0\Rightarrow a=3,4$  

 For a = 3, no solution

For a = 4, no solution

n (S₁) = 2, n (S₂) = 0,

Equation of intersection of line

$\frac{x-0}{1}=y=\frac{z-0}{-1}$

Let $\overrightarrow{r}=\widehat{i}-\widehat{k}$

Direction ratio of $\overrightarrow{PQ}$

$\Rightarrow \left(\lambda +1\right)\left(1\right)+\left(-2\right)\left(0\right)+\left(2-\lambda \right)\left(-1\right)=0$

$\lambda =\frac{1}{2}\Rightarrow Q\left(\frac{1}{2},0,\frac{-1}{2}\right)$

$PQ=\frac{\sqrt[]{34}}{2}$

$dist.=\frac{{\overrightarrow{A}}_1{A}_2.\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}$

$=\frac{\left(-4-\alpha ,-2,-3\right).\left(2,2,1\right)}{3}$

$\left|-5-\frac{2\alpha }{3}\right|=9\Rightarrow -5-\frac{2\alpha }{3}=\pm 9$

$\Rightarrow \frac{2\alpha }{3}=4,-14,$

but a > 0

a = 6

For the plane P,

$\overrightarrow{n}=\frac{1}{2}\left(-2\widehat{j}\right)*\left(-\widehat{i}+\widehat{j}-3\widehat{k}\right)=\widehat{j}*\widehat{i}+3\widehat{j}*\widehat{k}=-\widehat{k}+3\widehat{i}$

$\overrightarrow{a}.\overrightarrow{n}=0\Rightarrow 3\alpha -\gamma =0.......\left(i\right)$

$\overrightarrow{a}.\left(1,2,3\right)=0\Rightarrow \alpha +2\beta +3\gamma =0..........\left(ii\right)$

$\overrightarrow{a}.\left(1,1,2\right)=2\Rightarrow \alpha +\beta +2\gamma =2...........\left(iii\right)$

(i), (ii) & (iii) Þ a = 1, β = -5, $\gamma =$ 3

$T_{r+1}{=}^{120}{C}_r{4}^{\frac{120-r}{4}}.5^{r/6}$

For to be rational

r should be multiple of 6

Number of such terms = 21

$?\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=r\left(say\right)$

Let P (1 + 2r, 2 + 3r, 1 + 6r) lies on the plane 2x – y + z = 6

$\therefore r=1\Rightarrow P\left(3,5,5\right)$

Distance between P and Q is PQ =  $\sqrt[]{16+36+9}=\sqrt[]{61}$

$\therefore P{Q}^2=61$

 Normal vector to the given plane be

$2\widehat{i}-\widehat{j}+3\widehat{k}so$                   

Equation of line QS :

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

So let P $\left(2\lambda +1,-\lambda +3,\lambda +4\right)$  

Now P lies on given plane so

$4\lambda +2+\lambda -3+8\lambda +4+3=0$  

So, S (-3, 5, 2)

also given R lies on given plane so

6 – 5 + $\gamma$ + 3 = 0 so   $\gamma$ = -4

So, R (3, 5, -4)

SR² = 72

Required equation of plane will be (x – y – z – 1) + $\lambda$ (2x + y – 3z + 4) = 0

Given ${\perp }^r$ distance of (i) from origin = $\sqrt[]{\frac{2}{21}}$  

$\frac{\left|4\lambda -1\right|}{\sqrt[]{{\left(2\lambda +1\right)}^2+{\left(\lambda -1\right)}^2+{\left(3\lambda +1\right)}^2}}=\sqrt[]{\frac{2}{21}}$   

$\Rightarrow \lambda =\frac{1}{2}or\frac{15}{154}$

So plane be 4x – y – 5z + 2 = 0 for $\lambda =\frac{1}{2}$

Equation of line PQ :

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=k$

So, let Q (2k + 1, 3k – 2, -6k + 3) Q lies on given plane so

2k + 1 – 3k + 2 – 6k + 3 = 5

$-7k=-1ork=\frac{1}{7}$

So, $Q\left(\frac{9}{7},\frac{-11}{7},\frac{15}{7}\right)$

Now required distance = PQ = $\sqrt[]{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$

Equation of the plane will be $\left\{\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right\}+\lambda \left\{\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right\}=0$   

$\overrightarrow{r}.\left\{\left(1+2\lambda \right)i+\left(1+3\lambda \right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right\}+\left(4\lambda -1\right)=0$               

-> $\left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1-\lambda \right)z+\left(4\lambda -1\right)=0.........\left(i\right)$

(i) is parallel to x-axis so its d.r.s will be (1, 0, 0)

-> $1+2\lambda =0so\lambda =\frac{-1}{2}$

Hence required equation will be

$\overrightarrow{r}\left\{-\frac{1}{2}\widehat{j}.+\frac{3}{2}\widehat{k}\right\}+\left(-3\right)=0$

$\overrightarrow{r}.\left(\widehat{j}-3\widehat{k}\right)+6=0$