Plane is $a(x-3)+b(y-1)+c(z-1)=0$ 

Direction ratios of lines are$\left(1, -2, 2\right), \left(2, 3, -1\right)$.

Since both lines lie in to the plane implies dot product of direction ratios of line with direction ratio of normal of plane will be zero.

$\Rightarrow a-2b+2c=0$

$\Rightarrow 2a+3b-c=0$

$\frac{a}{-4}=\frac{b}{-\left(-5\right)}=\frac{c}{7}\Rightarrow \frac{{\displaystyle a}}{{\displaystyle 4}}=\frac{{\displaystyle b}}{{\displaystyle -5}}=\frac{{\displaystyle c}}{{\displaystyle -7}}$

$\therefore$ Direction ratios of normal of plane $\left(4,-5,-7\right)$ 

$\therefore$ Equation of plane is $4x-5y-7z=0$

 $\left(\alpha ,-3,5\right)$ lies on plane

$\therefore \alpha =5$