Given,

A plane passes through the points $(-1,k,0),(2,k,-1)$, $(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2y+1}{2}$ $=\frac{z+1}{-1}$,

So, on rearranging $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$ we get,

$\Rightarrow \frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$

Now given point : $A(-1,k,0),B(2,k,-1),C(1,1,2)$ 

So, $\overrightarrow{CA}=-2\hat{\mathit{i}}+(k-1)\hat{j}-2\hat{\mathit{k}}$, 

$\overrightarrow{CB}=\hat{\mathrm{i}}+(k-1)\hat{j}-3\hat{\mathit{k}}$

And $\overrightarrow{CA}\times \overrightarrow{CB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& k-1& -2\\ 1& k-1& -3\end{array}\right|$

$\Rightarrow \overrightarrow{CA}\times \overrightarrow{CB}=\hat{i}(-3k+3+2k-2)-\hat{j}(6+2)+\hat{k}(-2k+2-k+1)$

$\Rightarrow \overrightarrow{CA}\times \overrightarrow{CB}=\hat{i}(-3k+3+2k-2)-\hat{j}(6+2)+\hat{k}(-2k+2-k+1)$

$\Rightarrow \overrightarrow{CA}\times \overrightarrow{CB}=(1-k)\hat{i}-8\hat{j}+(3-3k)\hat{k}$

The line $\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$ is perpendicular to normal vector.

Now using the condition of perpendicular vector we get, 

$1(1-k)+1(-8)+(-1)(3-3k)=0$

$\Rightarrow 1-k-8-3+3k=0$

$\Rightarrow 2k=10\Rightarrow k=5$

Hence, $\frac{k^2+1}{(k-1)(k-2)}=\frac{26}{4\times 3}=\frac{13}{6}$