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

Answer 1

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

The family of planes passing through the intersection of given two planes is $x+y+z-1+\lambda (2x+3y-z+4)=0$
$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1-\lambda )z+4\lambda -1=0$
As the plane is parallel to $y$-axis,
$3\lambda +1=0\Rightarrow \lambda =-\frac{1}{3}$
So, the equation of the plane is
$(1-\frac{2}{3})x+(1+\frac{1}{3})z-\frac{7}{3}=0$
$\Rightarrow \frac{1}{3}x+\frac{4}{3}z-\frac{7}{3}=0\Rightarrow x+4z-7=0$

$\therefore$ The equation of plane is $\Rightarrow x+4z-7=0$
Hence the plane passes through point $(3,2,1)$