Let $P\left(2,-\mathrm{1,4}\right)$

Now, any point Q on line $\frac{x+3 }{10}=\frac{y-2}{-7}=\frac{z}{1}=\lambda$ is

$Q=\left(10\lambda -3, -7\lambda +2 ,\lambda \right)$

Direction Ratio of $PQ$ 

$=\left(10\lambda -3-2, -7\lambda +2+1, \lambda -4\right)$

$=\left(10\lambda -5, -7\lambda +3, \lambda -4\right)$

$\because PQ$ is perpendicular to given line

$\therefore 10\left(10\lambda -5\right)-7\left(-7\lambda +3\right)+ \left(\lambda -4\right)=0$

$\Rightarrow 150 \lambda =75 \Rightarrow \lambda =\frac{1}{2}$

$\therefore Q\equiv \left(2,-\frac{3}{2},\frac{1}{2}\right)$

Distance $PQ=\sqrt{\frac{1}{4}+\frac{49}{4}}=\frac{5}{\sqrt{2}}\approx 3.53$ which lies in $\left(\mathrm{3,4}\right)$.