Official solution
(3, 5, 7) lie on given line L1
\[{{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}\]
\[{{7 - b} \over 4} = 1 \Rightarrow b = 3\]
\[{{3 - a} \over l} = 1 \Rightarrow 3 - a = l\]
M(4, 3, 8)
N(3, 5, 7)
DR'S of MN = (1, -2, 1)
MN \[ \bot \] line L1
(1)(l) + (-2)(3) + 4(1) = 0
\[ \Rightarrow \] l = 2
a = 1
a = 1, b = 3, l = 2
\[{{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}\]
\[{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}\]
\[A = \, < 1,2,3 > \]
\[B = \, < 2,4,5 > \]
\[\overrightarrow {AB} = \, < 1,2,2 > \]
\[\overrightarrow p = 2\widehat i + 3\widehat j + 4\widehat k\]
\[\overrightarrow q = 3\widehat i + 4\widehat j + 5\widehat k\]
\[\overrightarrow p \, \times \overrightarrow q = - \widehat i + 2\widehat j - \widehat k\]
Shortest distance = \[\left| {{{\overrightarrow {AB} \,.\,(\overrightarrow p \times \overrightarrow q )} \over {|\overrightarrow p \times \overrightarrow q |}}} \right|\, = {1 \over {\sqrt 6 }}\]