The perpendicular distance of the point (2,4,-1) from line $\frac{\mathrm{x}+5}{1}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-6}{-9} \mathrm{is} \mathrm{\_\_\_\_\_\_\_}$

A = (2, 4, –1), Any point on the line P = (t – 5, 4t – 3, –9t + 6)

drs of AP=(t-7, 4t-7,-9t+7)

Let P be the foot of the perpendicular of A

AP is perpendicular to given line.

Use formula $$\begin{gathered} {\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2=0, \\ \Rightarrow 1(t-7)+4(4t-7)+(-9)(7-9t)=0 \\ \Rightarrow t+16t+81t-7-28-63=0 \\ \Rightarrow 98t-98=0 \\ \Rightarrow \mathrm{t}=1 \\ \therefore P(-4,1,-3) \\ AP=\sqrt{{(2+4)}^2+{(4-1)}^2+{(-1+3)}^2}=\sqrt{36+9+4}=7 \end{gathered}$$