If the shortest distance between the lines $\overrightarrow{\mathrm{r}}_{1}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}), \lambda \in \mathbf{R}, \alpha>0$ and $\overrightarrow{\mathrm{r}}_{2}=-4 \hat{\mathrm{i}}-\hat{\mathrm{k}}+\mu(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}), \mu \in \mathbf{R}$ is 9, then $\alpha$ is equal to
If $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{d}}$
then shortest distance between two lines is
$\mathrm{L}=\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}})}{|\mathrm{b} \times \mathrm{d}|}$
$\therefore \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}=((\alpha+4) \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
$\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}}{|\mathrm{b} \times \mathrm{d}|}=\frac{(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{3}$
$\therefore((\alpha+4) \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot \frac{(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{3}=9$
or $\alpha=6$
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