Official solution
Let \[{{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}\] = \[\lambda \]
Any arbitary point on the line is P( 2\[\lambda \] + 1, 3\[\lambda \] - 1, 4\[\lambda \] + 2).
This point also lies on the plane x + 2y + 3z = 15.
\[ \therefore \] (2\[\lambda \] + 1) + 2(3\[\lambda \] - 1) + 3(4\[\lambda \] + 2) = 15
\[ \Rightarrow \] 20\[\lambda \] = 10
\[ \Rightarrow \] \[\lambda \] = \[{1 \over 2}\]
So point P is (2, \[{1 \over 2}\], 4).
Distance from the origin(O) of point P(2, \[{1 \over 2}\], 4) is
OP = \[\sqrt {{{\left( 2 \right)}^2} + {{\left( {{1 \over 2}} \right)}^2} + {{\left( 4 \right)}^2}} \]
= \[\sqrt {4 + {1 \over 4} + 16} \]
= \[\sqrt {{{81} \over 4}} \]
= \[{9 \over 2}\]