<p>In throwing a dice thrice, getting numbers in order denoted by \(a, b, c\), satisfying \(a^2 + 4b^2 + 4c^2 - 2ab - 4bc - 2ac = 0\). If probability such that point \((a, b, c)\) lies inside the tetrahedron formed by the plane \(x + y + z = 10\) and co-ordinate planes is \(\dfrac{6}{\lambda}\), where \(\lambda \in N\), then \(\lambda\) is:</p>
Step-by-Step Solution
Key Concept: First, factor the constraint equation a² + 4b² + 4c² - 2ab - 4bc - 2ac = 0 as (a - b)² + (2b - 2c)² = 0, which forces a = b = 2c. Then identify which dice outcomes (a,b,c ∈ {1,2,3,4,5,6}) satisfy this relation and also lie inside the tetrahedron x + y + z < 10.
<p><strong>Step 1: Factor the constraint equation</strong></p><p>a² + 4b² + 4c² - 2ab - 4bc - 2ac = 0</p><p>Rearranging: a² - 2ab + b² + 3b² - 4bc + 4c² - 2ac = 0</p><p>= (a - b)² + (2b - 2c)² = 0</p><p>Since both squared terms are non-negative, each must equal zero:</p><p>(a - b)² = 0 and (2b - 2c)² = 0</p><p>Therefore: <strong>a = b = 2c</strong></p><p><strong>Step 2: Find valid dice outcomes</strong></p><p>With a, b, c ∈ {1, 2, 3, 4, 5, 6} and a = b = 2c:</p><p>• If c = 1: a = b = 2 ✓ (inside tetrahedron: 2 + 2 + 1 = 5 < 10)</p><p>• If c = 2: a = b = 4 ✓ (inside tetrahedron: 4 + 4 + 2 = 10, on boundary, excluded)</p><p>• If c = 3: a = b = 6 ✓ (inside tetrahedron: 6 + 6 + 3 = 15 > 10, outside)</p><p>• If c ≥ 4: a = 2c ≥ 8, impossible on dice</p><p><strong>Step 3: Count favorable outcomes</strong></p><p>Only (2, 2, 1) satisfies both constraints and lies strictly inside the tetrahedron.</p><p>Number of favorable outcomes = 1</p><p><strong>Step 4: Calculate probability</strong></p><p>Total possible outcomes when throwing dice thrice = 6³ = 216</p><p>Probability = 1/216 = 6/λ</p><p>Therefore: λ = 6 × 216 = <strong>1296</strong></p><p>∴ Answer: D (λ = 1296)</p>
Correct Answer: D