If the line

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

$\mathrm{x}=3 \lambda+2, \mathrm{y}=2 \lambda-1, \mathrm{z}=-\lambda+1$

Intersection with plane $2 x+3 y-z+13=0$

$2(3 \lambda+2)+3(2 \lambda-1)-(-\lambda+1)+13=0$

$13 \lambda+13=0 \quad \lambda=-1$

$\therefore \quad P(-1,-3,2)$

Intersection with plane

$3 x+y+4 z=16$

$3(3 \lambda+2)+(2 \lambda-1)+4(-\lambda+1)=16$

$\lambda=1$

$\mathrm{Q}(5,1,0)$

$\mathrm{PQ}=\sqrt{6^{2}+4^{2}+2^{2}}=\sqrt{56}=2 \sqrt{14}$