Question

If the distance of the point $(1,-2,3)$ from the plane $x+2y-3z+10=0$ measured parallel to the line, $\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}$ is $\sqrt{\frac{7}{2}},$ then the value of $|m|$ is equal to

Answer 1

$\frac{x-1}{3}=\frac{y+2}{-m}=\frac{z-3}{1}=\lambda$
Point $Q(3\lambda +1,-m\lambda -2,\lambda +3)$ lies on plane $x+2y-3z+10=0.$
$(3\lambda +1)+2(-m\lambda -2)-3(\lambda +3)+10=0$
$\Rightarrow 3\lambda -2m\lambda -3\lambda +1-4-9+10=0$
$\Rightarrow -2m\lambda =2$
$\Rightarrow m\lambda =-1\Rightarrow \lambda =-\frac{1}{m}$
So, coordinates of $Q$ are $(-\frac{3}{m}+1,-1,-\frac{1}{m}+3)$

Given, $PQ=\sqrt{\frac{7}{2}}$
$\Rightarrow \sqrt{{(-\frac{3}{m})}^2+1+{(-\frac{1}{m})}^2}=\sqrt{\frac{7}{2}}$
$\Rightarrow \frac{10+m^2}{m^2}=\frac{7}{2}$
$\Rightarrow 20+2m^2=7m^2$
$\Rightarrow m^2=4$
$\Rightarrow |m|=2$