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 find the value of  $(2d-8)$  is ______

Line through point  $P(-2,3,-4)$  and parallel to the given line
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  $4x+12y-3z+1=0$
$\Rightarrow$ Line is perpendicular to the normal to the plane 
 $$\begin{gathered} \Rightarrow 4\left(3\lambda \right)+12\left(\frac{4\lambda -9}{2}\right)-3\left(\frac{5\lambda +8}{3}\right)=0 \\ \Rightarrow \lambda =2\Rightarrow Q(4,\frac{5}{2},2) \\ \Rightarrow PQ=\sqrt{{\left(6\right)}^2+{(\frac{5}{2}-3)}^2+{\left(6\right)}^2}=\frac{17}{2} \end{gathered}$$