If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is :

<p>Let $P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)$ & $R(1,1,2)$</p> <p>$\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k$</p> <p>& $\overrightarrow Q R = - \widehat i + (1 - k)\widehat j + 3\widehat k$</p> <p>$\therefore$ Normal to plane will be</p> <p>$$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & {(1 - k)} & 2 \cr { - 1} & {(1 - k)} & 3 \cr } } \right| = \widehat i(1 - k) - \widehat j(8) + 3\widehat k(1 - k)$$</p> <p>If line is parallel to this we have</p> <p>$1(1 - k) + 1( - 8) + ( - 3)(1 - k) = 0$</p> <p>$\Rightarrow 2(1 - k) = - 8$</p> <p>$\Rightarrow 1 - k = - 4 \Rightarrow k = 5$</p> <p>$\therefore$ ${{{k^2} + 1} \over {(k - 1)(k - 2)}} = {{26} \over {4.3}} = {{13} \over 6}$</p>