Question

Let the image of the point $P(1,2,3)$ in the line $L:\frac{x-6}{3}=\frac{y-1}{2}=\frac{z-2}{3}$ 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

Answer 1

Answer: 125

Let $M$ be the mid-point of PQ
$\therefore M=(3\lambda +6,2\lambda +1,3\lambda +2)$
Now, $\overrightarrow{PM}=(3\lambda +5)\hat{i}+(2\lambda -1)\hat{j}+(3\lambda -1)\hat{k}$
$\because \overrightarrow{PM}\perp (3\hat{i}+2\hat{j}+3\hat{k})$
$\therefore 3(3\lambda +5)+2(2\lambda -1)+3(3\lambda -1)=0$
$\lambda =\frac{-5}{11}$
$\therefore M\left(\frac{51}{11},\frac{1}{11},\frac{7}{11}\right)$
Since R is mid-point of PM
$22(\alpha +\beta +\gamma )=125$