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…

Given \(P(2,-10,1)\)

\(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda\)

Let \(A\) is the point on the line,

\(A\left(2\lambda +1,-\lambda -2,2\lambda -3\right)\)

\(∵\vec{\mathrm{PA}}\cdot \vec{\mathrm{n}}=0\\ \Rightarrow \left(2\lambda -1\right)2+\left(-\lambda +8\right)\left(-1\right)+\left(2\lambda -4\right)2=0\\ \Rightarrow 4\lambda -2+\lambda -8+4\lambda -8=0\\ \Rightarrow 9\lambda -18=0\Rightarrow \lambda =2\\ ∴\mathrm{A}\left(5,-4,1\right)\\ ∴\mathrm{AP}=\sqrt{{3}^{2}+{6}^{2}+{0}^{2}}=\sqrt{45}=3\sqrt{5}\)