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

\[\begin{aligned} & l_{1}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}=\lambda \\ & \text { Any point on } l_{1} \text { is }(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & l_{2}: \frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-0}{1}=\mu \\ & \text { Any point on } l_{2} \text { is }(5 \mu+4,2 \mu+1, \mu) \\ & \text { For point of intersection, } \\ & 2 \lambda+1=5 \mu+4,3 \lambda+2=2 \mu+1 \\ & \text { Solving, } \lambda=\mu=-1 \\ & \text { Since, } \lambda=\mu=-1 \text { satisfy } 4 \lambda+3=\mu \\ & \therefore \text { Point of intersection is }(-1,-1,-1) \\ & \text { Now distance of }(-1,-5,-10) \text { from }(-1,-1,-1) \text { is: } \\ & \sqrt{(-1+1)^{2}+(-1+5)^{2}+(-1+10)^{2}}=\sqrt{97} \text { units } \end{aligned} \]