If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} \mathrm{k}$, and $\int_\limits 0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6 \alpha^3$ is equal to _________.

<p>$L_1: \frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$</p> <p>$$\begin{aligned} & \vec{b}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ & \vec{a}_1=-2 \hat{i}-3 \hat{j}+5 \hat{k} \end{aligned}$$</p> <p>$L_2=\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$</p> <p>$$\begin{aligned} & \vec{a}_2=3 \hat{i}+2 \hat{j}-4 \hat{k} \\ & \vec{b}_2=1 \hat{i}-3 \hat{j}+2 \hat{k} \end{aligned}$$</p> <p>$$d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|$$</p> <p>$$\begin{gathered} d=\left|\frac{(5 \hat{i}+5 \hat{j}-9 \hat{k}) \cdot(18 \hat{i}-9 \hat{k})}{\sqrt{324+81}}\right| \\ \left|\vec{b}_1 \times \vec{b}\right|\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2 \end{array}\right| \\ \Rightarrow \hat{i}(6+12)-\hat{j}(4-4)+\hat{k}(-6-3) \\ \Rightarrow(18 \hat{i}-9 \hat{k}) \end{gathered}$$</p> <p>$$\begin{aligned} & d=\left|\frac{90+81}{9 \sqrt{5}}\right| \\ & d=\frac{171}{9 \sqrt{5}} \\ & \frac{38}{3 \sqrt{5}} k=\frac{171}{9 \sqrt{5}} \\ & \frac{38}{3 \sqrt{5}} k=\frac{57}{3 \sqrt{5}} \end{aligned}$$</p> <p>$$\begin{aligned} & {k=\frac{57}{38}}=\frac{3}{2} \\ & \int_\limits0^{\frac{3}{2}}\left[x^2\right] d x \\ & \int_\limits0^1 0 d x+\int_\limits0^{\sqrt{ }} 1 d x+\int_\limits{\sqrt{2}}^{\frac{3}{2}} 2 d x \\ & 0+(\sqrt{2}-1)+2\left(\frac{3}{2}-\sqrt{2}\right) \\ & \sqrt{2}-1+3-2 \sqrt{2} \\ & 2-\sqrt{2} \end{aligned}$$</p> <p>$$\begin{aligned} & \alpha=2 \\ & 6 \alpha^3=6(2)^3=48 \end{aligned}$$</p>