Let the image of the point P ( 1, 2, 6 ) in the plane passing...

Plane passes through the points

A ( 1, 2, 0 )$,$B ( 1, 4, 1 ) and C ( 0, 5, 1 )$</p><p>We know that the equation of a plane passing through three non-collinear points$\left({x}_{1}, {y}_{1}, {z}_{1}\right)$, \left({x}_{2}, {y}_{2}, {z}_{2}\right)$, \left({x}_{3}, {y}_{3}, {z}_{3}\right)is \left|\begin{cases}x - {x}_{1} & y - {y}_{1} & z - {z}_{1} \\ {x}_{2} - {x}_{1} & {y}_{2} - {y}_{1} & {z}_{2} - {z}_{1} \\ {x}_{3} - {x}_{1} & {y}_{3} - {y}_{1} & {z}_{3} - {z}_{1}\end{cases}\right| = 0$</p><p>$⇒ \left|$\begin{cases}x - 1 & y - 2 & z - 0 \\ 1 - 1 & 4 - 2 & 1 - 0 \\ 0 - 1 & 5 - 2 & 1 - 0\end{cases}$\right|$= 0$

⇒ \left|\begin{cases}x - 1 & y - 2 & z - 0 \\ 0 & 2 & 1 \\ - 1 & 3 & 1\end{cases}\right| = \left(x - 1\right) $\left(2 - 3\right)$ - \left(y - 2\right) $\left(0 + 1\right)$ + z \left(0 + 2\right)$= 0$

$⇒ - x + 1 - y + 2 + 2z = 0$

$⇒ x + y – 2z = 3 . . . . \ldots ( i )$

Now image of $P ( 1, 2, 6 ) in ( i )$

\dfrac{x - 1}{1} = \dfrac{y - 2}{1} = \dfrac{z - 6}{- 2} = \dfrac{- 2 ( - 9 - 3 )}{6} =4$</p><p>$x = 5, y = 6, z = - 2$</p><p>$⇒ \alpha = 5, \beta = 6, \gamma = - 2$</p><p>$∴ {\alpha}^{2} + {\beta}^{2} + {\gamma}^{2} = 65

Hence this is the required option.