Question

Let the plane containing the line of intersection of the planes $P 1: x+(\lambda+4) y+z=1$ and $P 2: 2 x+$ $y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2 \lambda, \lambda,-\lambda)$ from the plane $P 2$ is

• A. $5 \sqrt{6}$
• B. $2 \sqrt{6}$
• C. $3 \sqrt{6}$
• D. $4 \sqrt{6}$

Answer 1

Answer: (C)

Equation of plane passing through point of intersection of $P 1$ and $P 2$
$ P=P 1+k P 2 $
$ (x+(\lambda+4) y+z-1)+k(2 x+y+z-2)=0$
Passing through $(0,1,0)$ and $(1,0,1)$
$ (\lambda+4-1)+ k (1-2)=0 $
$ (\lambda+3)- k =0 .....$(1)
Also passing $(1,0,1)$
$ (1+1-1)+ k (2+1-2)=0 $
$ 1+ k =0 $
$k =-1$
put in (1)
$ \lambda+3+1=0 $
$ \lambda=-4$
Then point $(2 \lambda, \lambda,-\lambda)$
$ d =\left|\frac{-16-4,-4,4)}{\sqrt{6}}\right| $
$ d =\frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=3 \sqrt{6}$