Let the image of the point $\mathrm{P}(1, 2, 6)$in the plane passing through the points $\mathrm{A}(1, 2, 0)$ and $\mathrm{B}(1, 4, 1) \mathrm{C}(0, 5, 1)$ be $\mathrm{Q}(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ . Then $\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2\right)$ equal to

Plane passes through the points

$\mathrm{A}(1, 2, 0)$, $\mathrm{B}(1, 4, 1)$ and $\mathrm{C}(0, 5, 1)$

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 $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-1& y-2& z-0\\ 1-1& 4-2& 1-0\\ 0-1& 5-2& 1-0\end{array}\right|=0$

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

$\Rightarrow -x+1-y+2+2z=0$

$\Rightarrow \mathrm{x} + \mathrm{y} – 2\mathrm{z} = 3 ....\dots \left(\mathrm{i}\right)$

Now image of $\mathrm{P}(1, 2, 6)$ in $\left(\mathrm{i}\right)$

$\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-6}{-2}=\frac{-2(-9-3)}{6}=4$

$x = 5, y = 6, z = -2$

$\Rightarrow \mathrm{\alpha }=5,\mathrm{\beta }=6,\mathrm{\gamma }=-2$

$\therefore {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2=65$

Hence this is the required option.