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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 $\int_0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6 \alpha^3$ is equal to________
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The correct answer is:
48
$\begin{aligned} & \frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{(5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-9 \hat{\mathrm{k}})}{\sqrt{5}} \cdot\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 4 \\ 1 & -3 & 2\end{array}\right| \\ & \frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{19}{\sqrt{5}} \\ & \mathrm{k}=\frac{19}{\sqrt{5}} \\ & \mathrm{k}=\frac{3}{2} \\ & \int_0^{3 / 2}\left[\mathrm{x}^2\right]=\int_0^1 0+\int_1^{\sqrt{2}} 1+\int_{\sqrt{2}}^{3 / 2} 2 \\ & =\sqrt{2}-1+2\left(\frac{3}{2}-\sqrt{2}\right) \\ & =2-\sqrt{2} \\ & \alpha=2 \\ & \Rightarrow 6 \alpha^3=48\end{aligned}$
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