If the mirror image of the point P ( 3, 4, 9 ) in the line (x -...

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

Let, $Q\left(\alpha, \beta, \gamma\right) be$ the image of P ( 3, 4, 9 ) and ' O '$be the mid-point of$PQwhich is on the $given line </p><p>\dfrac{x - 1}{3} = \dfrac{y + 1}{2} = \dfrac{z - 2}{1} =$\lambda$(say)</p><p>$⇒ x = 3\lambda + 1, y = 2\lambda - 1, z = \lambda + 2$</p><p>let \left(3\lambda + 1, 2\lambda - 1, \lambda + 2\right) be$ the co-ordinates of O$.</p><p>Direction ratios of the line is${a}_{1} = 3, {b}_{1} = 2, {c}_{1} = 1$</p><p>Also, direction ratios of$PO$are</p><p>\left(3\lambda - 2, 2\lambda - 5, \lambda - 7\right)</p><p>As,$PO$is perpendicular to the line</p><p>$3\left(3\lambda - 2\right)$+ 2\left(2\lambda - 5\right)$ + \lambda - 7 = 0$</p><p>$⇒ 14\lambda - 23 = 0$</p><p>$⇒ \lambda = $\dfrac{23}{14}$

$∴ O$ point is \left(3 \cdot \dfrac{23}{14} +1, 2 \cdot \dfrac{23}{14} -1, \dfrac{23}{14} +2\right)$</p><p>Since,$O$is mid-point of$PQ$</p><p>$∴ \left(3 \cdot \dfrac{23}{14} + 1, 2 \cdot \dfrac{23}{14} - 1$,$\dfrac{23}{14} + 2\right)$= \left(\dfrac{3 + \alpha}{2}, \dfrac{4 + \beta}{2}, \dfrac{9 + \gamma}{2}\right)$

So, on comparing both side and solving we get,

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

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