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 :

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Answer: D. 65 Equation of plane passing through the points A(1,2, 0), B(1,4,1) and C(0,5,1) is \begin{aligned} & \left|\begin{array}{ccc} x-1 & y-2 & z-0 \ 0 & 2 & 1 \ -1 & 3 & 1 \end{array}ight|=0 \\ & \Rightarrow x+y-2 z=3 \end{aligned} Now Q(\alpha, \beta, \gamma) is the image of the point P(1,2,6) in the plane x+y-2 z-3=0 \begin{array}{ll} &	herefore \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} 	ext { Hence, } \alpha^2+\beta^2+\gamma^2=5^2+6^2+(-2)^2=65

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 :

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