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

We seek the reflection $$Q(\alpha,\beta,\gamma)$$ of the point $$P(1,2,6)$$ in the plane determined by $$A(1,2,0)$$, $$B(1,4,1)$$ and $$C(0,5,1)$$, and then compute $$\alpha^2+\beta^2+\gamma^2$$.

First, the vectors in the plane are

$$\vec{AB}=B-A=(0,2,1),\quad\vec{AC}=C-A=(-1,3,1).$$

A normal vector follows from their cross product:

$$\vec{n}=\vec{AB}\times\vec{AC} =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&2&1\\-1&3&1\end{vmatrix} =\hat{i}(2-3)-\hat{j}(0+1)+\hat{k}(0+2)=(-1,-1,2).$$

Using $$A(1,2,0)$$, the plane equation becomes

$$-1(x-1)-1(y-2)+2(z-0)=0,$$ which simplifies to $$-x-y+2z+3=0\quad\text{or}\quad x+y-2z-3=0.$$

To reflect a general point $$(x_1,y_1,z_1)$$ in the plane $$ax+by+cz+d=0$$ one uses

$$(x-x_1)/a=(y-y_1)/b=(z-z_1)/c =\frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}.$$

Here $$a=1,\;b=1,\;c=-2,\;d=-3$$ and for $$P(1,2,6)$$ we compute

$$ax_1+by_1+cz_1+d=1+2-12-3=-12,\quad a^2+b^2+c^2=6,$$ so the common ratio is $$\frac{-2(-12)}6=4.$$

Hence

$$\alpha=1+1\cdot4=5,\quad\beta=2+1\cdot4=6,\quad\gamma=6+(-2)\cdot4=-2,$$

giving

$$Q=(5,6,-2).$$

Finally,

$$\alpha^2+\beta^2+\gamma^2=25+36+4=65.$$

The correct answer is Option 1: 65.