Let $$\vec{x}$$ be a vector in the plane containing vectors $$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$$. If the vector $$\vec{x}$$ is perpendicular to $$(3\hat{i} + 2\hat{j} - \hat{k})$$ and its projection on $$\vec{a}$$ is $$\frac{17\sqrt{6}}{2}$$, then the value of $$|\vec{x}|^2$$ is equal to ________.

The vector $$\vec{x}$$ lies in the plane of $$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$$, so we can write $$\vec{x} = \lambda\vec{a} + \mu\vec{b} = (2\lambda + \mu)\hat{i} + (-\lambda + 2\mu)\hat{j} + (\lambda - \mu)\hat{k}$$.

Since $$\vec{x}$$ is perpendicular to $$\vec{c} = 3\hat{i} + 2\hat{j} - \hat{k}$$, we have $$\vec{x} \cdot \vec{c} = 0$$: $$3(2\lambda + \mu) + 2(-\lambda + 2\mu) - (\lambda - \mu) = 0$$, giving $$6\lambda + 3\mu - 2\lambda + 4\mu - \lambda + \mu = 0$$, so $$3\lambda + 8\mu = 0$$, hence $$\lambda = -\frac{8\mu}{3}$$.

The projection of $$\vec{x}$$ on $$\vec{a}$$ is $$\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|} = \frac{17\sqrt{6}}{2}$$. We compute $$|\vec{a}| = \sqrt{4+1+1} = \sqrt{6}$$.

So $$\vec{x} \cdot \vec{a} = \frac{17\sqrt{6}}{2} \cdot \sqrt{6} = \frac{17 \cdot 6}{2} = 51$$.

Now $$\vec{x} \cdot \vec{a} = \lambda|\vec{a}|^2 + \mu(\vec{a} \cdot \vec{b})$$. We compute $$\vec{a} \cdot \vec{b} = 2 - 2 - 1 = -1$$ and $$|\vec{a}|^2 = 6$$. So $$6\lambda - \mu = 51$$.

Substituting $$\lambda = -\frac{8\mu}{3}$$: $$6 \cdot (-\frac{8\mu}{3}) - \mu = 51$$, giving $$-16\mu - \mu = 51$$, so $$-17\mu = 51$$, hence $$\mu = -3$$.

Then $$\lambda = -\frac{8(-3)}{3} = 8$$. So $$\vec{x} = (16 - 3)\hat{i} + (-8 - 6)\hat{j} + (8 + 3)\hat{k} = 13\hat{i} - 14\hat{j} + 11\hat{k}$$.

Therefore $$|\vec{x}|^2 = 169 + 196 + 121 = 486$$.