<p>If the point \((2, \alpha, \beta)\) lies on the plane which passes through the points \((3, 4, 2)\) and \((7, 0, 6)\) and is perpendicular to the plane \(2x - 5y = 15\), then \(2\alpha - 3\beta\) is equal to ______.</p>
Step-by-Step Solution
Key Concept: A plane perpendicular to another plane must have its normal vector perpendicular to the given plane's normal. Use this constraint along with the condition that two known points lie on the required plane to find its equation.
Step 1: Find the normal vector of the required plane. The given plane 2x - 5y = 15 has normal vector n_1 = (2, -5, 0). Since the required plane is perpendicular to this plane, its normal vector n must satisfy n · n_1 = 0. Also, vector through the two given points: v = (7-3, 0-4, 6-2) = (4, -4, 4). Let n = (a, b, c). Then: <ul><li>2a - 5b = 0 → a = (5b)/2</li><li> n · v = 0: 4a - 4b + 4c = 0 → a - b + c = 0</li></ul> Step 2: Solve for the normal vector. From a = (5b)/2 and a - b + c = 0: (5b)/2 - b + c = 0 → (3b)/2 + c = 0 → c = -(3b)/2 Let b = 2, then a = 5, c = -3. Normal vector n = (5, 2, -3). Step 3: Write the plane equation using point (3, 4, 2). 5(x - 3) + 2(y - 4) - 3(z - 2) = 0 5x + 2y - 3z = 15 + 8 - 6 = 17 Step 4: Substitute (2, α, β) into the plane equation. 5(2) + 2α - 3β = 17 10 + 2α - 3β = 17 2α - 3β = 7 ∴ Answer: 7
Correct Answer: 7