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 $(2d-8).$

Line through point $P(-2,3,-4)$ ad 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 palne. Thus,  $4\left(3\lambda \right)+12\left(\frac{4\lambda -9}{2}\right)-3\left(\frac{5\lambda +8}{3}\right)=0$
$Or \lambda =2 \Rightarrow Q\equiv (4,\frac{5}{2},2)$

$\Rightarrow PQ=\sqrt{{\left(6\right)}^2+{(\frac{5}{2}-3)}^2+{\left(6\right)}^2}=\frac{17}{2}$