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).

Line through point P(-2,3,4) from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ is $\frac{\mathrm{x}+2}{3}=\frac{\mathrm{y}+\frac{3}{2}}{2}=\frac{\mathrm{z}+\frac{4}{3}}{\frac{5}{3}}=\mathrm{\lambda }$

Any point on this line is $\mathrm{Q}\left[3\mathrm{\lambda }-2,2\mathrm{\lambda }-\frac{3}{2},\frac{5}{3}\mathrm{\lambda }-\frac{4}{3}\right]$

Direction ratios of PQ are $\left[3\lambda ,\frac{4\lambda -9}{2},\frac{5\lambda +8}{3}\right]$

Now PQ is parallel to the given plane  $4\mathrm{x}+12\mathrm{y}-3\mathrm{z}+1=0$

Hence, line is perpendicular to the normal to the plane. Thus, 

$\begin{array}{l}4\left(3\lambda \right)+12\left(\frac{4\lambda -9}{2}\right)-3\left(\frac{5\lambda +8}{3}\right)=0\\ \text{ or }\lambda =2\\ \Rightarrow Q\equiv \left(4,\frac{5}{2},2\right)\\ \Rightarrow PQ=\sqrt{(6{)}^2+{\left(\frac{5}{2}-3\right)}^2+(6{)}^2}=\frac{17}{2}\end{array}$