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

$$\begin{gathered} \mathrm{A}(1,2,0),\mathrm{B}(1,4,1),\mathrm{C}(0,5,1) \\ \text{ eq.plane is }\left|\begin{array}{ccc}\mathrm{x}-1& \mathrm{y}-2& \mathrm{z}\\ 0& 2& 1\\ -1& 3& 1\end{array}\right|=0 \\ \Rightarrow \mathrm{x}+\mathrm{y}-2\mathrm{z}-3=0 \\ \text{ image of }(1,2,6)\text{ in the plane }\mathrm{x}+\mathrm{y}-2\mathrm{z}-3=0\text{ is }(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }) \\ \begin{array}{l}\Rightarrow \frac{\mathrm{\alpha }-1}{1}=\frac{\mathrm{\beta }-2}{1}=\frac{\mathrm{\gamma }-6}{-2}=-2\left[\frac{1+2-12-3}{1^2+1^2+2^2}\right]\\ \frac{\alpha -1}{1}=\frac{\beta -2}{1}=\frac{\gamma -6}{-2}=4\\ \Rightarrow \mathrm{\alpha }=5,\mathrm{\beta }=6,\mathrm{\gamma }=-2\\ \therefore {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2=25+36+4=65\end{array} \end{gathered}$$