The length of perpendicular drawn from the point (3, –1,11) to the line $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$ is

Let $\mathrm{P}=(3,-1,11)$. Let Q be the foot of the perpendicular of P on the given line.

Let the coordinates of Q be $(2\mathrm{t}, 3\mathrm{t}+2, 4\mathrm{t}+3)$ which is any point on $\overrightarrow{\mathrm{AB}}$

D.r’s of $\mathrm{PQ}=(2\mathrm{t}-3, 3\mathrm{t}+3, 4\mathrm{t}-8)$

$\Rightarrow \text{ D'r's of }\mathrm{AB}=(2,3,4)$

AB is perpendicular to PQ

$\begin{array}{l}2(2\mathrm{t}-3)+3(3\mathrm{t}+3)+4(4\mathrm{t}-8)=0\\ \Rightarrow \mathrm{t}=1, \mathrm{Q}=(2,5,7), \mathrm{PQ}=\sqrt{53}\end{array}$