.If a plane passes through the point `(-1,k,0),(2,k,-1),(1,1,2)` and is parallel to the line `(x-1)/(1)=(2y+1)/(2)=(z+1)/(-1),` then the value of `(k^(2)+1)/((k-1)(k-2))` is

Options

A. `(13)/6`
B. `(5)/(17)`
C. `(17)/(5)`
D. `(6)/(13)`

Answer

To solve the problem, we need to find the value of \(\frac{k^2 + 1}{(k - 1)(k - 2)}\) given that a plane passes through the points \((-1, k, 0)\), \((2, k, -1)\), and \((1, 1, 2)\) and is parallel to the line defined by \(\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{-1}\).

### Step 1: Find the direction vector of the line
The line can be expressed in parametric form:
- \(x = 1 + t\)
- \(y = -\frac{1}{2} + \frac{t}{2}\)
- \(z = -1 - t\)

From this, we can extract the direction vector of the line:
\[ \vec{d} = (1, \frac{1}{2}, -1) \]

### Step 2: Determine the vectors from the points
Let \(A = (-1, k, 0)\), \(B = (2, k, -1)\), and \(C = (1, 1, 2)\).

Now, we compute the vectors \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{AB} = B - A = (2 - (-1), k - k, -1 - 0) = (3, 0, -1) \]
\[ \vec{AC} = C - A = (1 - (-1), 1 - k, 2 - 0) = (2, 1 - k, 2) \]

### Step 3: Find the normal vector of the plane
The normal vector \( \vec{n} \) to the plane can be found using the cross product \( \vec{AB} \times \vec{AC} \):
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 2 & 1 - k & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \left(0 \cdot 2 - (-1)(1 - k)\right) - \hat{j} \left(3 \cdot 2 - (-1)(2)\right) + \hat{k} \left(3(1 - k) - 0\right) \]
\[ = \hat{i} (1 - k) - \hat{j} (6 + 2) + \hat{k} (3 - 3k) \]
\[ = (1 - k, -8, 3 - 3k) \]

### Step 4: Set the condition for parallelism
Since the plane is parallel to the line, the normal vector \( \vec{n} \) must be perpendicular to the direction vector \( \vec{d} \). Thus, we set the dot product to zero:
\[ \vec{n} \cdot \vec{d} = (1 - k) \cdot 1 + (-8) \cdot \frac{1}{2} + (3 - 3k)(-1) = 0 \]
\[ (1 - k) - 4 + 3 - 3k = 0 \]
\[ 1 - k - 4 + 3 - 3k = 0 \]
\[ -4k + 0 = 0 \Rightarrow 2k - 2 = 0 \Rightarrow k = 5 \]

### Step 5: Substitute \(k\) into the expression
Now, substituting \(k = 5\) into the expression \(\frac{k^2 + 1}{(k - 1)(k - 2)}\):
\[ \frac{5^2 + 1}{(5 - 1)(5 - 2)} = \frac{25 + 1}{4 \cdot 3} = \frac{26}{12} = \frac{13}{6} \]

### Final Answer
Thus, the value of \(\frac{k^2 + 1}{(k - 1)(k - 2)}\) is \(\frac{13}{6}\).

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