Question

The distance of the point $(1,0,-3)$ from the plane $x-y-z=9$ measured parallel to the line $\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$ is

• A. $4$
• B. $8$
• C. $6$
• D. $7$

Answer 1

The correct option is D $7$

The equation of the line passing through $(1,0,-3)$ and parallel to the line $\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$ is
$\frac{x-1}{2}=$ $\frac{y-0}{3}=\frac{z+3}{-6}\cdots \left(i\right)$
which will intersect the plane $x-y-z=9\cdots \left(ii\right)$

Let any point on the line be $(2r+1,3r,-6r-3)$.
This point lies on the plane $\left(ii\right),$ Then

$(2r+1)-\left(3r\right)-(-6r-3)=9\Rightarrow 5r+4=9\Rightarrow r=1$

$\therefore$The point of intersection is $(3,3,-9)$. Now, the distance between the points $(1,0,-3)$ and $(3,3,-9)$
$=\sqrt{2^2+3^2+(-6{)}^2}=\sqrt{49}=7$