Key Concepts and Formulas

1. Vectors in a Plane: If three distinct points $A$, $B$, and $C$ lie on a plane, then the vectors $\vec{AB}$ and $\vec{AC}$ (or any other pair of vectors formed by these points) lie within that plane.
2. Normal Vector to a Plane ($\vec{N}$): A vector perpendicular to a plane can be found by taking the cross product of two non-collinear vectors lying in the plane. If $\vec{u}$ and $\vec{v}$ are in the plane, then $\vec{N} = \vec{u} \times \vec{v}$.
3. Direction Vector of a Line ($\vec{d}$): For a line in symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, its direction vector is $\vec{d} = \langle a, b, c \rangle$. It's crucial to ensure the coefficients of $x, y, z$ in the numerators are 1.
4. Condition for Parallelism between a Line and a Plane: A line is parallel to a plane if and only if its direction vector $\vec{d}$ is perpendicular (orthogonal) to the plane's normal vector $\vec{N}$. Mathematically, this means their dot product is zero: $\vec{d} \cdot \vec{N} = 0$.

Step-by-Step Solution

Step 1: Identify the Given Information We are given:

• Three points lying on the plane: $A = (-1, k, 0)$ $B = (2, k, -1)$ $C = (1, 1, 2)$
• A line parallel to the plane: $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$

Our objective is to calculate the value of the expression $\frac{k^2+1}{(k-1)(k-2)}$.

Step 2: Form Vectors Lying in the Plane To find the normal vector of the plane, we first construct two distinct vectors using the given points $A, B, C$. These vectors will lie within the plane.

Let's form $\vec{AB}$ and $\vec{AC}$: $\vec{AB} = B - A = (2 - (-1))\hat{i} + (k - k)\hat{j} + (-1 - 0)\hat{k}$ $\vec{AB} = 3\hat{i} + 0\hat{j} - 1\hat{k} = \langle 3, 0, -1 \rangle$ $\vec{AC} = C - A = (1 - (-1))\hat{i} + (1 - k)\hat{j} + (2 - 0)\hat{k}$ $\vec{AC} = 2\hat{i} + (1 - k)\hat{j} + 2\hat{k} = \langle 2, 1-k, 2 \rangle$ Reasoning: These vectors connect points on the plane, so they must also lie in the plane. They are non-collinear since their components are not proportional for any $k$, ensuring they define a unique plane.

Step 3: Determine the Normal Vector of the Plane ($\vec{N}$) The normal vector $\vec{N}$ to the plane is perpendicular to both $\vec{AB}$ and $\vec{AC}$. We obtain $\vec{N}$ by computing their cross product. $\vec{N} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 2 & 1-k & 2 \end{vmatrix}$ Expanding the determinant: $\vec{N} = \hat{i}((0)(2) - (-1)(1-k)) - \hat{j}((3)(2) - (-1)(2)) + \hat{k}((3)(1-k) - (0)(2))$ $\vec{N} = \hat{i}(0 + (1-k)) - \hat{j}(6 + 2) + \hat{k}(3 - 3k - 0)$ $\vec{N} = (1-k)\hat{i} - 8\hat{j} + (3-3k)\hat{k} = \langle 1-k, -8, 3(1-k) \rangle$ Reasoning: The cross product of two vectors in a plane yields a vector that is orthogonal to both, and thus normal to the plane containing them.

Step 4: Determine the Direction Vector of the Line ($\vec{d}$) The given equation of the line is $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$. To find the direction vector, we must ensure the coefficients of $x, y, z$ in the numerators are 1. The $y$-term needs adjustment: $\frac{2 y+1}{2} = \frac{2(y + 1/2)}{2} = \frac{y + 1/2}{1}$ So, the standard symmetric form of the line equation is: $\frac{x-1}{1}=\frac{y + 1/2}{1}=\frac{z+1}{-1}$ From this, the direction vector of the line is $\vec{d} = \langle 1, 1, -1 \rangle$. Reasoning: The direction ratios of a line are the denominators in its standard symmetric equation. It's crucial to first convert the given equation to the standard form.

**Step 5: Apply the Parallelism Condition and Solve for $k$} Since the plane is parallel to the line, their normal vector $\vec{N}$ and the line's direction vector $\vec{d}$ must be perpendicular. Therefore, their dot product must be zero. $\vec{N} \cdot \vec{d} = 0$ Substitute the components of $\vec{N} = \langle 1-k, -8, 3(1-k) \rangle$ and $\vec{d} = \langle 1, 1, -1 \rangle$: $(1-k)(1) + (-8)(1) + (3(1-k))(-1) = 0$ $1 - k - 8 - (3 - 3k) = 0$ $1 - k - 8 - 3 + 3k = 0$ $2k - 10 = 0$ $2k = 10$ $k = 5$ Reasoning: The dot product of two perpendicular vectors is zero. This condition allows us to set up an equation and solve for the unknown parameter $k$.

Step 6: Calculate the Required Expression Now that we have found $k=5$, we can substitute this value into the expression $\frac{k^2+1}{(k-1)(k-2)}$. $\frac{k^2+1}{(k-1)(k-2)} = \frac{(5)^2+1}{(5-1)(5-2)}$ $= \frac{25+1}{(4)(3)}$ $= \frac{26}{12}$ $= \frac{13}{6}$

Common Mistakes & Tips

• Standard Form of Line Equation: Always convert the line equation to the standard symmetric form ($\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$) before identifying the direction vector. Failing to do so for terms like $2y+1$ is a common error.
• Cross Product Calculation: Be meticulous with signs and calculations when expanding the determinant for the cross product. A single sign error can lead to an incorrect normal vector.
• Parallelism vs. Perpendicularity: Clearly distinguish between the conditions for a line being parallel to a plane ($\vec{d} \cdot \vec{N} = 0$) and a line being perpendicular to a plane ($\vec{d}$ is parallel to $\vec{N}$, i.e., $\vec{d} = \lambda \vec{N}$).

Summary

We first found two vectors lying in the plane using the given points. Their cross product yielded the normal vector to the plane in terms of $k$. We then determined the direction vector of the given line by converting its equation to standard form. By applying the condition that a line parallel to a plane has its direction vector perpendicular to the plane's normal vector (i.e., their dot product is zero), we formed an equation for $k$. Solving this equation gave $k=5$. Finally, substituting $k=5$ into the required expression $\frac{k^2+1}{(k-1)(k-2)}$ yielded $\frac{13}{6}$.

The final answer is $\boxed{\text{(C) } \frac{13}{6}}$.