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

$$\begin{gathered} \mathrm{P}(3,-1,11) \\ given line \frac{\mathrm{x}}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}=t \\ \Rightarrow Foot of the perpendicular drawn from P to the \\ line=\mathrm{Q}(2\mathrm{t},3\mathrm{t}+2,4\mathrm{t}+3) \\ \therefore d{r}^{\mathrm{\text{'}}}\mathrm{s} \mathrm{of} \mathrm{PQ}=(2\mathrm{t}-3,3\mathrm{t}+3,4\mathrm{t}-8) \\ the line PQ is perpendicular to the given line \\ \Rightarrow 2(2\mathrm{t}-3)+3(3\mathrm{t}+3)+4(4\mathrm{t}-8)=0 \\ \Rightarrow \mathrm{t}=1 \\ \Rightarrow \mathrm{P}(3,-1,11)\mathrm{Q}(2,5,7) \\ \therefore \mathrm{PQ}=\sqrt{1+36+16} =\sqrt{53} \end{gathered}$$