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

$$\begin{gathered} \left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]=\overrightarrow{u}\cdot \left(\overrightarrow{v}\times \overrightarrow{w}\right) \\ min.\left(\left|u\right|\left|\overrightarrow{v}\times \overrightarrow{w}\right|\cos \theta \right)=-\alpha \sqrt{3401} \end{gathered}$$
$$\begin{gathered} \Rightarrow \cos \theta =-1 \\ \begin{array}{r}\left|u\right|=\alpha \left(\text{ Given }\right)\\ |\overrightarrow{v}\times \overrightarrow{w}|=\sqrt{3401}\\ \overrightarrow{v}\times \overrightarrow{w}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ \alpha & 2& -3\\ 2\alpha & 1& -1\end{array}\right|\\ \overrightarrow{v}\times \overrightarrow{w}=\widehat{i}-5\alpha \widehat{j}-3\alpha \widehat{k} \\ \end{array} \end{gathered}$$
$\begin{array}{l}|\overrightarrow{v}\times \overrightarrow{w}|=\sqrt{1+25{\alpha }^2+9{\alpha }^2}=\sqrt{3401}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}34{\alpha }^2=3400\\ {\alpha }^2=100\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{(as }\alpha >0)\end{array}$
$$\begin{gathered} \begin{array}{l}so, \\ \overrightarrow{u}=\lambda (\widehat{i}-5\alpha \widehat{j}-3\alpha \widehat{k})\\ \overrightarrow{u}=\sqrt{{\lambda }^2+25{\alpha }^2{\lambda }^2+9{\alpha }^2\lambda }\\ {\alpha }^2={\lambda }^2\left(1+25{\alpha }^2+9{\alpha }^2\right)\\ 100={\lambda }^2(1+34\times 100)\\ {\lambda }^2=\frac{100}{3401}=\frac{m}{n}\end{array} \end{gathered}$$