Let the image of the point P(1, 2, 3) in the line $L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}$ be Q. Let R ($\alpha$, $\beta$, $\gamma$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($\alpha$ + $\beta$ + $\gamma$) is equal to __________.

<p>The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.</p> <p>$\therefore$ Let a point on line be $\lambda$</p> <p>$\Rightarrow {{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3} = \lambda$</p> <p>$\Rightarrow P'(3\lambda + 6,\,2\lambda + 1,\,3\lambda + 2)$</p> <p>as P' is foot of perpendicular</p> <p>$(3\lambda + 5)3 + (2\lambda - 1)2 + (3\lambda - 1)3 = 0$</p> <p>$\Rightarrow 22\lambda + 15 - 2 - 3 = 0$</p> <p>$\Rightarrow \lambda = {{ - 5} \over {11}}$</p> <p>$\therefore$ $P'\left( {{{51} \over {11}},{1 \over {11}},{7 \over {11}}} \right)$</p> <p>Mid-point of $$PP' \equiv \left( {{{{{51} \over {11}} + 1} \over 2},{{{1 \over {11}} + 2} \over 2},{{{7 \over {11}} + 3} \over 2}} \right)$$</p> <p>$$ \equiv \left( {{{62} \over {22}},{{23} \over {22}},{{40} \over {22}}} \right) \equiv (\alpha ,\beta ,\gamma )$$</p> <p>$\Rightarrow 22(\alpha ,\beta ,\gamma ) = 62 + 23 + 40 = 125$</p>