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 :
Correct Option: , 3
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)
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