Let three vectors $$\vec{a} = \alpha\hat{i} + 4\hat{j} + 2\hat{k}$$, $$\vec{b} = 5\hat{i} + 3\hat{j} + 4\hat{k}$$, $$\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$$ form a triangle such that $$\vec{c} = \vec{a} - \vec{b}$$ and the area of the triangle is $$5\sqrt{6}$$. If $$\alpha$$ is a positive real number, then $$|\vec{c}|^2$$ is equal to:

$$\vec{c} = \vec{a} - \vec{b} = (\alpha-5)\hat{i} + \hat{j} - 2\hat{k}$$.

Area of triangle = $$\frac{1}{2}|\vec{a} \times \vec{b}| = 5\sqrt{6}$$, so $$|\vec{a} \times \vec{b}| = 10\sqrt{6}$$.

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix} = (16-6)\hat{i} - (4\alpha-10)\hat{j} + (3\alpha-20)\hat{k}$$

$$= 10\hat{i} - (4\alpha-10)\hat{j} + (3\alpha-20)\hat{k}$$

$$|\vec{a} \times \vec{b}|^2 = 100 + (4\alpha-10)^2 + (3\alpha-20)^2 = 600$$

$$(4\alpha-10)^2 + (3\alpha-20)^2 = 500$$

$$16\alpha^2 - 80\alpha + 100 + 9\alpha^2 - 120\alpha + 400 = 500$$

$$25\alpha^2 - 200\alpha + 500 = 500$$

$$25\alpha^2 - 200\alpha = 0 \implies 25\alpha(\alpha - 8) = 0$$

Since $$\alpha > 0$$: $$\alpha = 8$$.

$$\vec{c} = (8-5)\hat{i} + \hat{j} - 2\hat{k} = 3\hat{i} + \hat{j} - 2\hat{k}$$.

$$|\vec{c}|^2 = 9 + 1 + 4 = 14$$.

The correct answer is Option 2: 14.