If the shortest distance between the lines $\overline{\mathrm{r}}_1=\alpha \hat{i}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+\lambda(\hat{i}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}), \lambda \in \mathbb{R}, \alpha>0 \text { and }$ $\overline{\mathrm{r}}_2=-4 \hat{i}-\hat{\mathrm{k}}+\mu(3 \hat{i}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}), \mu \in \mathrm{R}$, is 9, then the value of $\alpha$ is

Given the skew lines:

$\vec{r}_1 = (\alpha\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$

$\vec{r}_2 = (-4\hat{i} - \hat{k}) + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})$

The shortest distance between skew lines is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

The direction vectors are $\vec{b}_1 = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b}_2 = 3\hat{i} - 2\hat{j} - 2\hat{k}$.

The cross product evaluates to $\vec{b}_1 \times \vec{b}_2 = 8\hat{i} + 8\hat{j} + 4\hat{k}$ with magnitude 12.

The difference between position vectors is $\vec{a}_2 - \vec{a}_1 = (-4 - \alpha)\hat{i} - 2\hat{j} - 3\hat{k}$.

The dot product simplifies to $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = -8\alpha - 60$.

Setting the distance equal to 9 gives $\frac{|-8\alpha - 60|}{12} = 9$, leading to $|8\alpha + 60| = 108$.

This yields $\alpha = -21$ or $\alpha = 6$. Since $\alpha > 0$, the appropriate value is $\alpha = 6$.