If a plane passes through the points (-1, k, 0), (2, k, -1), (1, 1, 2) and is parallel to the line x−11=2y+12=z+1−1, then the value of k2+1(k−1)(k−2) is

Equation of plane is x−1y−1z−211−130−1=0⇒x + 2y + 3z = 9(-1, k, 0) is on the plane ⇒k = 5k2+1(k−1)(k−2)=264×3=136

If a plane passes through the points (-1, k, 0), (2, k, -1), (1, 1, 2) and is parallel to the line x−11=2y+12=z+1−1, then the value of k2+1(k−1)(k−2) is

Equation of plane is x−1y−1z−211−130−1=0⇒x + 2y + 3z = 9(-1, k, 0) is on the plane ⇒k = 5k2+1(k−1)(k−2)=264×3=136

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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

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5/17

13/6

6/13

17/5

answer is D.

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Detailed Solution

Equation of plane is $|\begin{array}{ccc} x - 1 & y - 1 & z - 2 \\ 1 & 1 & - 1 \\ 3 & 0 & - 1 \end{array}| = 0$
$\Rightarrow$ x + 2y + 3z = 9
(-1, k, 0) is on the plane $\Rightarrow$ k = 5
$\frac{k^{2} + 1}{(k - 1) (k - 2)} = \frac{26}{4 \times 3} = \frac{13}{6}$