If the lines $\overrightarrow{r}=(\hat{i}-\hat{j}+\hat{k})+\lambda (3\hat{j}-\hat{k})$ and $\overrightarrow{r}=(\alpha \hat{i}-\hat{j})+\mu (2\hat{i}-3\hat{k})$ are co-planar, then distance of the plane containing these two lines from the point $(\alpha ,0,0)$ is :

$$\begin{gathered} \overrightarrow{r}=(\hat{i}-\hat{j}+\hat{k})+\lambda (3\hat{j}-\hat{k}) .....\left(1\right) \\ \overrightarrow{r}=(\alpha \hat{i}-\hat{j})+\mu (2\hat{i}-3\hat{k}) .....\left(2 \right) \end{gathered}$$

(1) and (2) are coplanar

$$\begin{gathered} \therefore \left|\begin{array}{ccc}0& 3& -1\\ 2& 0& -3\\ (1-\alpha )& 0& 1\end{array}\right|=0 \\ -3(2+3(1-\alpha \left)\right)=0 \\ 2+3-3\alpha =0 \\ 3\alpha =5 \end{gathered}$$

Now,

$$\begin{gathered} \overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 3& -1\\ 2& 0& -3\end{array}\right|=\hat{i}(-9)-\hat{j}\left(2\right)+k(-6) \\ =(9,2,6) \end{gathered}$$

Equation of plane:

9(x-1)+2(y+1)+6(z-1)=0

9 x+2 y+6 z-13=0

Perpendicular distance from $(5/3,0,0)$

$=\left|\frac{\left(9·\frac{5}{3}+0+0-13\right)}{\sqrt{81+36+4}}\right|=\frac{2}{\sqrt{121}}=\frac{2}{11}$