We have,

$\overrightarrow{r_1}=\alpha \hat{i}+2\hat{j}+2\hat{k}+\lambda \left(\hat{i}-2\hat{j}+2\hat{k}\right)$

$\overrightarrow{r_2}=-4\hat{i}-\hat{k}+\mu \left(3\hat{i}-2\hat{j}-2\hat{k}\right)$

If $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ and $\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{d}$, then shortest distance between two lines is

$L=\frac{\left(\overrightarrow{a}-\overrightarrow{c}\right)·\left(\overrightarrow{b}\times \overrightarrow{d}\right)}{\left|\overrightarrow{b}\times \overrightarrow{d}\right|}$

Here,

$\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 2\\ 3& -2& -2\end{array}\right|$

$\Rightarrow \overrightarrow{b}\times \overrightarrow{d}=8\hat{i}+8\hat{j}+4\hat{k}$

$\Rightarrow \overrightarrow{b}\times \overrightarrow{d}=4\left(2\hat{i}+2\hat{j}+\hat{k}\right)$

$\Rightarrow \left|\overrightarrow{b}\times \overrightarrow{d}\right|=4·3$

And,

$\overrightarrow{a}-\overrightarrow{c}=\left(\left(\alpha +4\right)\hat{i}+2\hat{j}+3\hat{k}\right)$

So, shortest distance is

$\frac{\left(\left(\alpha +4\right)\hat{i}+2\hat{j}+3\hat{k}\right)·4\left(2\hat{i}+2\hat{j}+\hat{k}\right)}{4·3}=9$

$\Rightarrow 2\left(\alpha +4\right)+4+3=27$

$\Rightarrow \left(\alpha +4\right)=10$

$\Rightarrow \alpha =6$