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

Equation of plane passing through the points $A(1,2$, $0), B(1,4,1)$ and $C(0,5,1)$ is <br/><br/>$$ \begin{aligned} & \left|\begin{array}{ccc} x-1 & y-2 & z-0 \\ 0 & 2 & 1 \\ -1 & 3 & 1 \end{array}\right|=0 \\\\ & \Rightarrow x+y-2 z=3 \end{aligned} $$ <br/><br/>Now $Q(\alpha, \beta, \gamma)$ is the image of the point $P(1,2,6)$ in the plane $x+y-2 z-3=0$ <br/><br/>$$ \begin{array}{ll} &\therefore \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2[1+2-2(6)-3]}{1^2+1^2+(-2)^2} \\\\ &\Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4 \\\\ &\Rightarrow \alpha=5, \beta=6, \gamma=-2 \end{array} $$ <br/><br/>$\text { Hence, } \alpha^2+\beta^2+\gamma^2=5^2+6^2+(-2)^2=65$