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

Answer 1

The correct option is B $(3,2,1)$

Given: $p_1:x+y+z=1$ and $p_2:2x+3y-z+4=0$
$\Rightarrow p_1+\lambda p_2=0$
$\Rightarrow (x+y+z-1)+\lambda (2x+3y-z+4)=0\cdots \left(i\right)$
$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1-\lambda )z-1+4\lambda =0$
dr's of normal of the plane are $1+2\lambda ,1+3\lambda ,1-\lambda$
Since plane is parallel to y-axis, $1+3\lambda =0$
$\lambda =-\frac{1}{3}$
putting in equation $\left(i\right)$
$\Rightarrow (x+y+z-1)+\frac{-1}{3}(2x+3y-z+4)=0$
$\Rightarrow x+4z-7=0$
option $2$ $(3,2,1)$ satisfies the equation $=3+4-7=7-7=0$