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

Given,

$\overrightarrow{\mathrm{r}}=\left(\hat{i}-\hat{j}+\hat{k}\right)+\lambda \left(3\hat{j}-\hat{k}\right) ...L1$

$\overrightarrow{\mathrm{r}}=\left(\alpha \hat{i}-\hat{j}\right)+\mu \left(2\hat{i}-3\hat{k}\right)$

$L1$ and $L2$ are coplanar then $\left[\begin{array}{ccc}{a}_2-a_1& b_1& b_2\end{array}\right]=0$

So, $\left|\begin{array}{ccc}0& 3& -1\\ 2& 0& -3\\ \left(1-\alpha \right)& 0& 1\end{array}\right|=0$

$-3\left(2+3\left(1-\alpha \right)\right)=0$

$3\alpha =5$

$\Rightarrow \alpha =\frac{5}{3}$
Now, normal vector will be, $\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 3& -1\\ 2& 0& -3\end{array}\right|=\hat{i}\left(-9\right)-\hat{j}\left(2\right)+k\left(-6\right)$

$=\left(9,2,6\right)$

So, equation of plane is given by, 

$9\left(\mathrm{x}-1\right)+2\left(\mathrm{y}+1\right)+6\left(\mathrm{z}-1\right)=0$

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

Now finding perpendicular distance from $\left(\alpha ,0,0\right)$

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