If a plane passes through the points ( - 1, k, 0 ) , ( 2, k, - 1 )...

Given,

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

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

$⇒ \dfrac{x - 1}{1} = \dfrac{y + \dfrac{1}{2}}{1} = \dfrac{z + 1}{- 1}$

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

So, \overset{\to}{CA} = - 2\overset{^}{i} + ( k - 1 ) \overset{^}{j} - 2\overset{^}{k},

\overset{\to}{CB} = \overset{^}{i} + ( k - 1 ) \overset{^}{j} - 3\overset{^}{k}

And \overset{\to}{CA} \times \overset{\to}{CB} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 2 & k - 1 & - 2 \\ 1 & k - 1 & - 3\end{cases}\right|

⇒ \overset{\to}{CA} \times \overset{\to}{CB} = \overset{^}{i} ( - 3k + 3 + 2k - 2 ) - \overset{^}{j} ( 6 + 2 ) + \overset{^}{k} ( - 2k + 2 - k + 1 )

⇒ \overset{\to}{CA} \times \overset{\to}{CB} = \overset{^}{i} ( - 3k + 3 + 2k - 2 ) - \overset{^}{j} ( 6 + 2 ) + \overset{^}{k} ( - 2k + 2 - k + 1 )

⇒ \overset{\to}{CA} \times \overset{\to}{CB} = ( 1 - k ) \overset{^}{i} - 8\overset{^}{j} + ( 3 - 3k ) \overset{^}{k}

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

Now using the condition of perpendicular vector we get,

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

$⇒ 1 - k - 8 - 3 + 3k = 0$

$⇒ 2k = 10 ⇒ k = 5$

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