Question

The distance of the point $P(-2,3,-4)$ from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ measured parallel to the plane $4x+12y-3z+1=0$ is $d$, then the value of $(2d-8)$ is

Answer 1

Line through point $P(-2,3,-4)$ and parallel to the given line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ is $\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}=\lambda$
Any point on this line is $Q[3\lambda -2,2\lambda -\frac{3}{2},\frac{5}{3}\lambda -\frac{4}{3}]$

Direction ratios of $PQ$ are $[3\lambda ,\frac{4\lambda -9}{2},\frac{5\lambda +8}{3}]$
Now $PQ$ is parallel to the given plane $4x+12y-3z+1=0$

Hence, line is perpendicular to the normal to the plane.
Thus, $4\left(3\lambda \right)+12\left(\frac{4\lambda -9}{2}\right)-3\left(\frac{5\lambda +8}{3}\right)=0$
or $\lambda =2$
$\Rightarrow Q(4,\frac{5}{2},2)$
$\Rightarrow PQ=\sqrt{(6{)}^2+{(\frac{5}{2}-3)}^2+(6{)}^2}=\frac{17}{2}$