Question

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

• A. 70
• B. 72
• C. 74
• D. 76

Answer 1

The Correct Option is C

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: \( \alpha\beta + \beta\gamma + \gamma\alpha = 74 \).