Let two vertices of triangle ABC be (2, 4, 6) and (0, –2, –5), and its centroid be (2, 1, –1). If the image of third vertex in the plane x + 2y + 4z = 11 is (α, β, γ), then αβ + βγ + γα is equal to
- 70
- 72
- 74
- 76
The Correct Option is C
Solution and Explanation
Given: Two vertices of the triangle are \( A(2, 4, 6) \) and \( B(0, -2, -5) \), and the centroid is \( G(2, 1, -1) \). Let the third vertex be \( C(x, y, z) \).
- Step 1: Use the centroid formula:
\( \frac{2 + 0 + x}{3} = 2, \quad \frac{4 - 2 + y}{3} = 1, \quad \frac{6 - 5 + z}{3} = -1 \).
- Solve for \( x, y, z \):
- \( \frac{2 + 0 + x}{3} = 2 \implies x = 4 \).
- \( \frac{4 - 2 + y}{3} = 1 \implies y = 1 \).
- \( \frac{6 - 5 + z}{3} = -1 \implies z = -4 \).
- Step 2: Find the image of vertex \( C(4, 1, -4) \) in the plane \( x + 2y + 4z - 11 = 0 \). Let the image be \( D(\alpha, \beta, \gamma) \). The formula for the reflection is:
\( \frac{\alpha - 4}{1} = \frac{\beta - 1}{2} = \frac{\gamma + 4}{4} = \frac{-4 + 2 - 16 - 11}{1 + 4 + 16} \).
- Simplify:
\( \frac{\alpha - 4}{1} = \frac{\beta - 1}{2} = \frac{\gamma + 4}{4} = \frac{-25}{21} \).
- Solve for \( \alpha, \beta, \gamma \):
- \( \alpha - 4 = \frac{-25}{21} \implies \alpha = 6 \).
- \( \beta - 1 = 2 \cdot \frac{-25}{21} \implies \beta = 5 \).
- \( \gamma + 4 = 4 \cdot \frac{-25}{21} \implies \gamma = 4 \).
- Step 3: Calculate \( \alpha\beta + \beta\gamma + \gamma\alpha \):
\( \alpha\beta + \beta\gamma + \gamma\alpha = (6 \cdot 5) + (5 \cdot 4) + (4 \cdot 6) \).
- Simplify:
\( 30 + 20 + 24 = 74 \).
Final Answer: \( \alpha\beta + \beta\gamma + \gamma\alpha = 74 \).
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