Official solution
Equation of AB,
\[{{x - 1} \over 0} = {{y + 2} \over 3} = {{z - 3} \over { - 3}} = \lambda \]
\[ \therefore \] Coordinates of any point on the line (M) = \[\left( { - 3,3\lambda - 2, - 3\lambda } \right)\]
\[\overrightarrow {PM} = - 3\widehat i + \left( {3\lambda - 4} \right)\widehat j - 3\lambda \widehat k\]
\[\overrightarrow {AB} = 3\widehat j - 3\widehat k\]
As \[\overrightarrow {PM} \bot \overrightarrow {AB} \]
\[ \therefore \] \[\overrightarrow {PM} .\overrightarrow {AB} = 0\]
\[ \Rightarrow \] \[\left( { - 3} \right).0 + \left( {3\lambda - 4} \right)\left( 3 \right) + \left( { - 3\lambda } \right)\left( { - 3} \right)\] = 0
\[ \Rightarrow \] \[\lambda \] = \[{2 \over 3}\]
\[ \therefore \] M = (1, 0, 1)
By checking each options we can see M lies on 2x + y -- z = 1.