Point (3λ + 1, λ + 2, 2λ + 3) on first line lies on second line.

$\therefore \frac{3\mathrm{\lambda }+1-3}{1}=\frac{\mathrm{\lambda }+2-1}{2}=\frac{2\mathrm{\lambda }+3-2}{3}$

⇒ 6λ – 4 = λ + 1 ⇒ λ = 1

$\therefore$ Point of intersection = (4, 3, 5)

$\because$Equation of plane P will be

4 (x – 4) + 3 (y – 3) + 5 (z – 5) = 0

⇒ 4x +3y + 5z = 50.

Let a point B lie on line

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

$\therefore$ B = (k + 1, k – 2, 2k + 3)

$\because$ AB is parallel to plane.

$\therefore$ AB is normal to normal of the plane.

$\therefore \overrightarrow{\mathrm{AB}}=\mathrm{k}\hat{\mathrm{i}}+(\mathrm{k}-4)\hat{\mathrm{j}}+(2\mathrm{k}-1)\hat{\mathrm{k}}\mathrm{and}\mathrm{normal}\overrightarrow{\mathrm{n}}=4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+5\hat{\mathrm{k}}$

4k + 3(k – 4) + 10k – 5 = 0 

⇒ 17k = 17 ⇒ k = 1

$\therefore$ Point B = (2, – 1, 5)

$\therefore \mathrm{AB}=\sqrt{1^2+3^2+1^2}=\sqrt{11}.$