The length of the perpendicular from the point (2, -1, 4) on the straight line, $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$  is:

$\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}=\lambda$ 

$\begin{array}{c}\left(10\lambda -3,-7\lambda +2,\lambda \right)\\ \left(2,-1,4\right)\end{array}$

Dr’s          $\left(10\lambda -5,-7\lambda +3,\lambda -4\right)$

$\begin{array}{c}10(10\lambda -5)-7(-7\lambda +3)+1(\lambda -4)=0\\ \lambda =\frac{1}{2}\\ \text{perpendicular distance = }\sqrt{\frac{50}{4}}\text{∈}(3,4)\end{array}$