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 $$[\vec{u} \quad \vec{v} \quad \vec{w}]$$ 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 _____.

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

First, compute $$\vec{v}\times\vec{w}$$ by evaluating the determinant $$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\alpha&2&-3\\2\alpha&1&-1\end{vmatrix}$$, yielding $$\hat{i}(1)-\hat{j}(5\alpha)+\hat{k}(-3\alpha)=\langle1,-5\alpha,-3\alpha\rangle$$.

By Cauchy-Schwarz, the minimum of $$\vec{u}\cdot\vec{n}$$ subject to $$|\vec{u}|=\alpha$$ is $$-\alpha|\vec{n}|$$.

Since $$|\vec{n}|=\sqrt{1+25\alpha^2+9\alpha^2}=\sqrt{1+34\alpha^2}$$, setting $$-\alpha\sqrt{1+34\alpha^2}=-\alpha\sqrt{3401}$$ gives $$1+34\alpha^2=3401$$, so $$\alpha^2=100$$ and $$\alpha=10$$.

The minimum occurs when $$\vec{u}=-\frac{\alpha}{|\vec{n}|}\vec{n}$$, hence $$\vec{u}\cdot\hat{i}=-\frac{10}{\sqrt{3401}}\cdot1$$, which implies $$|\vec{u}\cdot\hat{i}|^2=\frac{100}{3401}$$.

Since $$\gcd(100,3401)=1$$ (as $$3401=19\times179$$), we set $$m=100$$ and $$n=3401$$, yielding

$$m+n=\boxed{3501}$$.