Question

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

• A. $(-3,0,1)$
• B. $(3,3,-1)$
• C. $(3,2,1)$
• D. $(-3,1,1)$

Answer 1

Answer: (C)

$(3,2,1)$

Equation of plane

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

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

dr's of normal of the plane are

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

Since plane is parallel to y-axis,

$1+3\lambda =0$

$\Rightarrow \lambda =\frac{-1}{3}$

So the equation of plane is

$x+4z-7=0$

Point $(3,2,1)$ satisfies this equation

Hence Answer is $\left(3\right)$.