Question

The plane containing the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}$ and also containing its projection on the plane $2x+3y-z=5$, contains which one of the following points ?

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

Answer 1

The correct option is B $(2,0,-2)$

The required plane passes through $(3,-2,1)$
Let DR's of required plane be $(a,b,c)$
$\therefore a(x-3)+b(y+2)+c(z-1)=0\dots \left(1\right)$
Also, $2a-b+3c=0\dots \left(2\right)$
$2a+3b-c=0\dots \left(3\right)$
From $\left(2\right)$ and $\left(3\right),$
$a:b:c=-1:1:1$
$\therefore -(x-3)+y+2+z-1=0$
$\Rightarrow -x+y+z+4=0$
or, $x-y-z-4=0$