To find what \(\cos^2\theta\) is equal to, we need to use the coordinates provided for points \(A\), \(B\), and \(C\) and apply some vector geometry and trigonometry concepts.

Step 1: Understand the Geometry of the Problem

We are given the points as:

• \(A (1, -1, 0)\)
• \(B(-2, 1, 8)\)
• \(C(-1, 2, 7)\)

These points are consecutive vertices of a parallelogram \(ABCD\).

Step 2: Find Vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \)

Let's calculate the vectors:

• \(\overrightarrow{AB} = B - A = (-2 - 1, 1 - (-1), 8 - 0) = (-3, 2, 8)\)
• \(\overrightarrow{BC} = C - B = (-1 - (-2), 2 - 1, 7 - 8) = (1, 1, -1)\)

Step 3: Calculate the Dot Product and Magnitudes

We need to find the angle \(\theta\) between vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \).

Dot Product:

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = (-3)(1) + (2)(1) + (8)(-1) = -3 + 2 - 8 = -9\)

Magnitude of Vectors:

• \(|\overrightarrow{AB}| = \sqrt{(-3)^2 + 2^2 + 8^2} = \sqrt{9 + 4 + 64} = \sqrt{77}\)
• \(|\overrightarrow{BC}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}\)

Step 4: Find \(\cos \theta\) and \(\cos^2 \theta \)

The cosine of the angle \(\theta\) is given by:

\(\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{BC}}{|\overrightarrow{AB}| |\overrightarrow{BC}|} = \frac{-9}{\sqrt{77} \times \sqrt{3}}\)

\(\Rightarrow \cos \theta = \frac{-9}{\sqrt{231}}\)

Thus, \(\cos^2 \theta\) is:

\(\cos^2 \theta = \left(\frac{-9}{\sqrt{231}}\right)^2 = \frac{81}{231} = \frac{27}{77}\)

Conclusion

Therefore, the value of \(\cos^2 \theta\) is \(\frac{27}{77}\), corresponding to the correct answer option.

Correct Answer: \(27/77\)