The plane through the intersection of the planes
Question:

The plane through the intersection of the planes $\mathrm{x}+\mathrm{y}+\mathrm{z}=1$ and $2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}+4=0$ and parallel to $y$-axis also passes through the point :

1. $(-3,0,-1)$

2. $(3,3,-1)$

3. $(3,2,1)$

4. $(-3,1,1)$

Correct Option: , 3

Solution:

Equation of plane

$(x+y+z-1)+\lambda(2 x+3 y-z+4)=0$

$\Rightarrow(1+2 \lambda) \mathrm{x}+(1+3 \lambda) \mathrm{y}+(1-\lambda) \mathrm{z}-1+4 \lambda=0$

dr’s of normal of the plane are

$1+2 \lambda, 1+3 \lambda, 1-\lambda$

Since plane is parallel to $\mathrm{y}-$ axis, $1+3 \lambda=0$

$\Rightarrow \quad \lambda=-1 / 3$

So the equation of plane is

$x+4 z-7=0$

Point $(3,2,1)$ satisfies this equation Hence Answer is (3)