Question

A plane $P$ contains the line $x+2y+3z+1=0=x-y-z-6,$ and is perpendicular to the plane $-2x+y+z+8=0.$ Then which of the following points lies on $P?$

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

Answer 1

The correct option is C $(0,1,1)$

Required plane is a plane passing through the line of intersection of planes
$P_1\equiv x+2y+3z+1=0$
and $P_2\equiv x-y-z-6=0$
Its equation: $P_1+\lambda P_2=0$
$\Rightarrow (x+2y+3z+1)+\lambda (x-y-z-6)=0$
$\Rightarrow (1+\lambda )x+(2-\lambda )y+(3-\lambda )z+1-6\lambda =0$
$\because$ Perpendicular to $-2x+y+z+8=0$
$\therefore -2(1+\lambda )+(2-\lambda )+(3-\lambda )=0$
$\Rightarrow \lambda =\frac{3}{4}$
$\Rightarrow$ Required plane is $7x+5y+9z=14$

Checking the option shows that $(0,1,1)$ satisfies it.