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)

$3 \sqrt{6}$

Answer 2

The correct answer is (C) : $3 \sqrt{6}$
Equation of plane passing through point of intersection of P1 and P2
P=P1+kP2
(x+(λ+4)y+z−1)+k(2x+y+z−2)=0
Passing through (0,1,0) and (1,0,1)
(λ+4−1)+k(1−2)=0
(λ+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)
λ+3+1=0
λ=−4
Then point (2λ,λ,−λ)
d=∣∣​6​−16−4,−4,4)​∣∣​
d=6​18​×6​6​​=36​