The line which contains all points (x, y, z) which are of the form $x\hat{i}+y\hat{j}+z\hat{k}=\left(2\hat{i}-2\hat{j}+5\hat{k}\right)+\lambda \left(\hat{i}-3\hat{j}+2\hat{k}\right)$ intersects the plane 2x – 3y + 4z = 163 , at P and intersects the YZ plane at Q. If the distance PQ is $a\sqrt{b}$  where    then the value of -----------

Equation of the line is $\frac{x-2}{1}=\frac{y+2}{-3}=\frac{z-5}{2}=\lambda$....(1)
 hence any point on the line (1) can be taken as
   
 for some   point lies on the plane
    ....(2)
    
 This gives 
                     
 Hence, P   (9, –23, 19)
 Also (1) intersect YZ plane  i.e.  x = 0   + 2 = 0,  hence   = – 2
   Q(0, 4, 1)
 So, $PQ=\sqrt{9^2+{27}^2+{18}^2}=9 \sqrt{1+3^2+2^2}=9\sqrt{14}$

   a = 9 and b = 14
 Hence, a + b = 23