A plane passing through the point (3,1,1)

Hence normal is $\perp^{\mathrm{r}}$ to both the lines so normal vector to the plane is

$\overrightarrow{\mathrm{n}}=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})$

$\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -2 & 2 \\ 2 & 3 & -1\end{array}\right|=\hat{\mathrm{i}}(2-6)-\hat{\mathrm{j}}(-1-4)+\hat{\mathrm{k}}(3+4)$

$\overrightarrow{\mathrm{n}}=-4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

Now equation of plane passing through $(3,1,1)$ is

$\Rightarrow-4(x-3)+5(y-1)+7(z-1)=0$

$\Rightarrow-4 x+12+5 y-5+7 z-7=0$

$\Rightarrow-4 x+5 y+7 z=0$......(1)

Plane is also passing through $(\alpha,-3,5)$ so this point satisfies the equation of plane so put in equation (1)

$-4 \alpha+5 \times(-3)+7 \times(5)=0$

$\Rightarrow-4 \alpha-15+35=0$

$\Rightarrow \alpha=5$