Let $$P$$ be a plane passing through the points $$(1, 0, 1)$$, $$(1, -2, 1)$$ and $$(0, 1, -2)$$. Let a vector $$\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$ be such that $$\vec{a}$$ is parallel to the plane $$P$$, perpendicular to $$(\hat{i} + 2\hat{j} + 3\hat{k})$$ and $$\vec{a} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 2$$, then $$(\alpha - \beta + \gamma)^2$$ equals ___.

We first find the equation of plane $$P$$ through points $$(1,0,1)$$, $$(1,-2,1)$$, and $$(0,1,-2)$$. Two vectors in the plane are $$\vec{v_1} = (0,-2,0)$$ and $$\vec{v_2} = (-1,1,-3)$$.

The normal to the plane is $$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&-2&0\\-1&1&-3\end{vmatrix} = (6, 0, -2)$$, simplified to $$(3, 0, -1)$$.

The plane equation is $$3x - z = 2$$.

Now let $$\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$. The three conditions are:

(i) $$\vec{a}$$ is parallel to plane $$P$$: $$\vec{a} \cdot \vec{n} = 0 \implies 3\alpha - \gamma = 0 \implies \gamma = 3\alpha$$.

(ii) $$\vec{a}$$ is perpendicular to $$\hat{i}+2\hat{j}+3\hat{k}$$: $$\alpha + 2\beta + 3\gamma = 0 \implies \alpha + 2\beta + 9\alpha = 0 \implies \beta = -5\alpha$$.

(iii) $$\vec{a} \cdot (\hat{i}+\hat{j}+2\hat{k}) = 2$$: $$\alpha + \beta + 2\gamma = 2 \implies \alpha - 5\alpha + 6\alpha = 2 \implies 2\alpha = 2 \implies \alpha = 1$$.

So $$\alpha = 1$$, $$\beta = -5$$, $$\gamma = 3$$, and:

$$(\alpha - \beta + \gamma)^2 = (1 - (-5) + 3)^2 = (9)^2 = 81$$