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

Let a point on the given line be
$Q(3k-2,\frac{4k-3}{2},\frac{5k-4}{3})$
Direction ratios of $PQ$ will be
$(3k,\frac{4k-9}{2},\frac{5k+8}{3})$
As we know $PQ$ will be parallel to the plane, so it will be perpendicular to its normal.
$PQ$ is parallel to the given plane.
$\Rightarrow 4\times 3k+12\left(\frac{4k-9}{2}\right)-3\left(\frac{5k+8}{3}\right)=0\Rightarrow k=2$
So, the coordinates of $Q$ is $(4,\frac{5}{2},2)$
So, $PQ=\sqrt{36+\frac{1}{4}+36}=\frac{17}{2}=d$
$\Rightarrow 2d-8=9$