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

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

Determine the direction vector of the given line.

Rewriting the line equation: $$\frac{x-1}{1} = \frac{y + 1/2}{1} = \frac{z+1}{-1}$$

The direction vector is $$\vec{d} = (1, 1, -1)$$.

Find two vectors lying in the plane.

$$\vec{AB} = B - A = (2-(-1), k-k, -1-0) = (3, 0, -1)$$

$$\vec{AC} = C - A = (1-(-1), 1-k, 2-0) = (2, 1-k, 2)$$

Compute the normal vector to the plane.

$$\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 2 & 1-k & 2 \end{vmatrix}$$

$$= \hat{i}(0 \cdot 2 - (-1)(1-k)) - \hat{j}(3 \cdot 2 - (-1) \cdot 2) + \hat{k}(3(1-k) - 0 \cdot 2)$$

$$= (1-k)\hat{i} - 8\hat{j} + 3(1-k)\hat{k}$$

Apply the condition that the plane is parallel to the line.

If the plane is parallel to the line, then the normal to the plane is perpendicular to the direction of the line:

$$\vec{n} \cdot \vec{d} = 0$$

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

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

$$2k - 10 = 0 \implies k = 5$$

Compute the required expression.

$$\frac{k^2 + 1}{(k-1)(k-2)} = \frac{25 + 1}{(5-1)(5-2)} = \frac{26}{4 \times 3} = \frac{26}{12} = \frac{13}{6}$$

The correct answer is Option (4): $$\frac{13}{6}$$.