Since the plane contain two lines whose direction ratios are $\left(-1,2,1\right)$ and $\left(1,3,2\right)$ .

So, normal vector to the plane is given by $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 2& 1\\ 1& 3& 2\end{array}\right|=\hat{i}+3\hat{j}-5\hat{k}$

Since, the plane passes through $\left(2,1,k\right)$, its equation is given by

$\left(x-2\right)+3\left(y-1\right)-5\left(z-k\right)=0$

If it passes through $\left(3,-1,4\right)$, then 

$\left(3-2\right)+3\left(-1-1\right)-5\left(4-k\right)=0\Rightarrow k=5$