The length of the perpendicular from the point \(\mathrm{A}(1,-2,-3)\) on the line \(\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}\) is

Let \(\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}=\lambda\)
\(\therefore \quad\) Any general point on this line is
\(\mathrm{Q}(2 \lambda+1,-\lambda-3,-2 \lambda-1)\)
The direction ratios of AQ are
\(2 \lambda,-\lambda-1,-2 \lambda+2\)
Since \(A Q\) is perpendicular to given lines.
\(\begin{aligned}
& 2(2 \lambda)-1(-\lambda-1)-2(-2 \lambda+2)=0 \\
\Rightarrow & 4 \lambda+\lambda+1+4 \lambda-4=0 \\
& \Rightarrow 9 \lambda-3=0 \\
\Rightarrow & \lambda=\frac{1}{3} \\
\therefore \quad & \mathrm{Q} \equiv\left(\frac{5}{3}, \frac{-10}{3}, \frac{-5}{3}\right)
\end{aligned}\)
\(\therefore \quad \mathrm{AQ}=\sqrt{\left(1-\frac{5}{3}\right)^2-\left(-2+\frac{10}{3}\right)^2+\left(-3+\frac{5}{3}\right)^2}\)
\(\begin{aligned} & =\sqrt{\frac{4}{9}+\frac{16}{9}+\frac{16}{9}}=\sqrt{\frac{36}{9}}=\sqrt{4} \\ \therefore \quad A Q & =2 \text { units }\end{aligned}\)