The distance of the point (-1, 9, -16) from the plane 2x + 3y - z = 5 measured parallel to the line $\frac{\mathrm{x}+4}{3}=\frac{2-\mathrm{y}}{4}=\frac{\mathrm{z}-3}{12}$ is

Distance of $\left(-1,9,-16\right)$  from plane 2x + 3y -z - 5 = 0
Measured parallel to line $\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-2}{-4}=\frac{\mathrm{z}-3}{12}$
 

Is $\left|\frac{{\mathrm{ax}}_1+{\mathrm{by}}_1+{\mathrm{cz}}_1+\mathrm{d}}{\mathrm{al}+\mathrm{bm}+\mathrm{cn}}\right|$ where l, m, n are d.c’ s of line

$\left|\frac{2\left(-1\right)+3\left(9\right)-1\left(-16\right)-5}{2\left({\displaystyle \frac{3}{13}}\right)+3\left({\displaystyle \frac{-4}{13}}\right)-1\left({\displaystyle \frac{12}{13}}\right)}\right|=\left|\frac{36\times 13}{-18}\right|$= 26