Thank you for reporting, we will resolve it shortly
Q.
If a plane pasees through the pointe $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is
$ \frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1} $
$ \frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1} $
$ \text { Points : } A(-1, k , 0), B (2, k ,-1), C (1,1,2) $
$ \overrightarrow{ CA }=-2 \hat{ i }+( k -1) \hat{ j }-2 \hat{ k } $
$\overrightarrow{ CB }=\hat{ i }+( k -1) \hat{ j }-3 \hat{ k }$
$\overrightarrow{ CA } \times \overrightarrow{ CB }=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & k -1 & -2 \\ 1 & k -1 & -3\end{vmatrix}$
$=\hat{ i }(-3 k +3+2 k -2)-\hat{ j }(6+2)+\hat{ k }(-2 k +2- k +1)$
$ =(1- k ) \hat{ i }-8 \hat{ j }+(3-3 k ) \hat{ k }$
The line $\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$ is perpendicular to normal vector.
$ \therefore 1 \cdot(1- k )+1(-8)+(-1)(3-3 k )=0$
$ \Rightarrow 1- k -8-3+3 k =0 $
$ \Rightarrow 2 k =10 \Rightarrow k =5 $
$ \therefore \frac{ k ^2+1}{( k -1)( k -2)}=\frac{26}{4 \cdot 3}=\frac{13}{6}$