Find the foot of the perpendicular drawn from point $\displaystyle (2,-1,5)$ to the line $\displaystyle \frac{\mathrm{x}-11}{10}=\frac{\mathrm{y}+2}{-4}=\frac{\mathrm{z}+8}{-11}$. Also, find the length of the perpendicular.

Let I: $\displaystyle \frac{\mathrm{x}-11}{10}=\frac{\mathrm{y}+2}{-4}=\frac{\mathrm{z}+8}{-11}=\lambda$ Coordinates of any point on /are $\displaystyle \mathrm{x}=10 \lambda+11, \mathrm{y}=-4 \lambda-2, \mathrm{z}=-11 \lambda-8$ Drs of perpendicular line are $\displaystyle (10 \lambda+9,-4 \lambda-1,-11 \lambda-13)$ Drs of given line are $\displaystyle 10$, -$\displaystyle 4$,-$\displaystyle 11$ As lines are perpendicular, so \[\begin{aligned} & (10 \lambda+9) 10+(-4 \lambda-1)(-4)+(-11 \lambda-13) \times(-11)=0 \\ & \Rightarrow \lambda=-1 \end{aligned} \] Hence coordinates of point are ( $\displaystyle 1,2,3$ ) which is the foot of the ⟂ from $\displaystyle P$ to $\displaystyle l$. length of $\displaystyle \perp=\sqrt{(\mathbf{1}-\mathbf{2})^{\mathbf{2}}+(\mathbf{2}+\mathbf{1})^{\mathbf{2}}+(\mathbf{3}-\mathbf{5})^{\mathbf{2}}}=\sqrt{\mathbf{1}+\mathbf{9}+\mathbf{4}}=\sqrt{\mathbf{1 4}}$