The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.

$$\therefore$$ Let a point on line be $$\lambda$$

$$ \Rightarrow {{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3} = \lambda $$

$$ \Rightarrow P'(3\lambda + 6,\,2\lambda + 1,\,3\lambda + 2)$$

as P' is foot of perpendicular

$$(3\lambda + 5)3 + (2\lambda - 1)2 + (3\lambda - 1)3 = 0$$

$$ \Rightarrow 22\lambda + 15 - 2 - 3 = 0$$

$$ \Rightarrow \lambda = {{ - 5} \over {11}}$$

$$\therefore$$ $$P'\left( {{{51} \over {11}},{1 \over {11}},{7 \over {11}}} \right)$$

Mid-point of $$PP' \equiv \left( {{{{{51} \over {11}} + 1} \over 2},{{{1 \over {11}} + 2} \over 2},{{{7 \over {11}} + 3} \over 2}} \right)$$

$$ \equiv \left( {{{62} \over {22}},{{23} \over {22}},{{40} \over {22}}} \right) \equiv (\alpha ,\beta ,\gamma )$$

$$ \Rightarrow 22(\alpha ,\beta ,\gamma ) = 62 + 23 + 40 = 125$$