The correct option is B 2√14
Given line x−23=y+12=z−1−1= λ(let)
Therefore general point of line:
(3λ+2, 2λ−1, 1−λ) which intersect the plane 2x+3y−z+13=0 at a point P
then 2(3λ+2)+3(2λ−1)−(1−λ)+13=0⇒λ=−1
So, point P(−1,−3, 2)
Also line intersect the plane 3x+y+4z=16 at a point Q
therefore 3(3λ+2)+(2λ−1)+4(1−λ)=16⇒λ=1
So, point Q(5,1,0)
therefore PQ=√(5+1)2+(1+3)2+(0−2)2
∴PQ=2√14