Let $$\vec{v} = \alpha\hat{i} + 2\hat{j} - 3\hat{k}$$, $$\vec{w} = 2\alpha\hat{i} + \hat{j} - \hat{k}$$, and $$\vec{u}$$ be a vector such that $$|\vec{u}| = \alpha > 0$$. If the minimum value of the scalar triple product $$[\vec{u}\ \vec{v}\ \vec{w}]$$ is $$-\alpha\sqrt{3401}$$, and $$|\vec{u} \cdot \hat{i}|^2 = \frac{m}{n}$$ where $$m$$ and $$n$$ are coprime natural numbers, then $$m + n$$ is equal to _____.

Given: $$\vec{v} = \alpha\hat{i} + 2\hat{j} - 3\hat{k}$$, $$\vec{w} = 2\alpha\hat{i} + \hat{j} - \hat{k}$$, $$|\vec{u}| = \alpha > 0$$.

The scalar triple product $$[\vec{u}\ \vec{v}\ \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$$.

First, compute $$\vec{v} \times \vec{w}$$.

$$ \vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2\alpha & 1 & -1 \end{vmatrix} = \hat{i}(-2+3) - \hat{j}(-\alpha+6\alpha) + \hat{k}(\alpha-4\alpha) $$ $$ = \hat{i}(1) - \hat{j}(5\alpha) + \hat{k}(-3\alpha) = (1, -5\alpha, -3\alpha) $$

Next, find minimum of $$[\vec{u}\ \vec{v}\ \vec{w}]$$.

The minimum value of $$\vec{u} \cdot (\vec{v} \times \vec{w})$$ with $$|\vec{u}| = \alpha$$ is $$-\alpha|\vec{v} \times \vec{w}|$$, occurring when $$\vec{u}$$ is antiparallel to $$\vec{v} \times \vec{w}$$.

$$ |\vec{v} \times \vec{w}| = \sqrt{1 + 25\alpha^2 + 9\alpha^2} = \sqrt{1 + 34\alpha^2} $$

Given: minimum = $$-\alpha\sqrt{3401}$$

$$ -\alpha\sqrt{1 + 34\alpha^2} = -\alpha\sqrt{3401} $$ $$ 1 + 34\alpha^2 = 3401 $$ $$ \alpha^2 = 100, \quad \alpha = 10 $$

Now, find $$|\vec{u} \cdot \hat{i}|^2$$.

When the minimum occurs, $$\vec{u} = -\frac{\alpha}{|\vec{v} \times \vec{w}|}(\vec{v} \times \vec{w}) = -\frac{10}{\sqrt{3401}}(1, -50, -30)$$

$$ \vec{u} \cdot \hat{i} = -\frac{10}{\sqrt{3401}} $$ $$ |\vec{u} \cdot \hat{i}|^2 = \frac{100}{3401} $$

Since $$\gcd(100, 3401) = 1$$ (as $$3401 = 34 \times 100 + 1$$), we have $$m = 100$$, $$n = 3401$$.

$$ m + n = 100 + 3401 = 3501 $$

Therefore, $$m + n = 3501$$.