The perpendicular distance, of the line \(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}\) from the point \(P(2,-10,1)\), is…

\(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda \\ \text{Let}A\text{is the point on the line,}\\ A(2\lambda +1,-\lambda -2,2\lambda -3)\text{and}P(2,-10,1)\\ ∵\vec{\mathrm{PA}}\cdot \vec{\mathrm{n}}=0\\ \Rightarrow (2\lambda -1)2+(-\lambda +8)(-1)+(2\lambda -4)2=0\\ \Rightarrow 4\lambda -2+\lambda -8+4\lambda -8=0\\ \Rightarrow 9\lambda -18=0\Rightarrow \lambda =2\\ ∴\mathrm{A}(5,-4,1)\\ ∴\mathrm{AP}=\sqrt{{3}^{2}+{6}^{2}+{0}^{2}}=\sqrt{45}=3\sqrt{5}\)