Find the distance of the point (-$\displaystyle 1$, -$\displaystyle 5$, -$\displaystyle 10$) from the point of intersection of the line $\displaystyle \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}$ and the plane $\displaystyle \mathrm{x}-\mathrm{y}+\mathrm{z}=5$.

General point of the line is \[(3 \lambda+2,4 \lambda-1,12 \lambda+2) \] This will be the point of intersection with the plane, if \[\begin{aligned} & (3 \lambda+2)-(4 \lambda-1)+(12 \lambda+2)=5 \\ \Rightarrow & 11 \lambda+5=5, \Rightarrow \lambda=0 \end{aligned} \] Thus, the point of intersection is ($\displaystyle 2$, -$\displaystyle 1$, $\displaystyle 2$) Distance \(\displaystyle =\sqrt{(2+1)^{2}+(-1+5)^{2}+(2+10)^{2}}\) \[\begin{aligned} & =\sqrt{9+16+144} \\ & =\sqrt{169}=13 \end{aligned} \]