Mathematics - Vector Question with Solution | TestHub

The given point is $P(3,4,9)$.

Let, $Q\left(\alpha ,\beta ,\gamma \right)$ be the image of $P(3,4,9)$ and $\text{'}O\text{'}$ be the mid-point of $PQ$ which is on the given line 

$\frac{x-1}{3}=\frac{{\displaystyle y+1}}{{\displaystyle 2}}=\frac{{\displaystyle z-2}}{{\displaystyle 1}}=\lambda$ (say)

$\Rightarrow x=3\lambda +1, y=2\lambda -1, z=\lambda +2$ 

let $\left(3\lambda +1, 2\lambda -1, \lambda +2\right)$ be the co-ordinates of $O$.

Direction ratios of the line is $a_1=3, b_1=2, c_1=1$

Also, direction ratios of $PO$ are

$\left(3\lambda -2, 2\lambda -5, \lambda -7\right)$

As, $PO$ is perpendicular to the line

$3\left(3\lambda -2\right)+2\left(2\lambda -5\right)+\lambda -7=0$

$\Rightarrow 14\lambda -23=0$

$\Rightarrow \lambda =\frac{23}{14}$

$\therefore$ $O$ point is $\left(3·\frac{23}{14}+1, 2·\frac{23}{14}-1, \frac{23}{14}+2\right)$

Since, $O$ is mid-point of $PQ$

$\therefore \left(3·\frac{23}{14}+1, 2·\frac{23}{14}-1, \frac{23}{14}+2\right)=\left(\frac{3+\alpha }{2},\frac{4+\beta }{2},\frac{9+\gamma }{2}\right)$

So, on comparing both side and solving we get,

$\alpha =\frac{62}{7}, \beta =\frac{4}{7}, \gamma =\frac{-12}{7}$

Hence, $14\left(\alpha + \beta +\gamma \right)=108$