Question:

<p><strong>Step 1:</strong> Rewrite the equation in standard form. Given: 4x² - 4xy + ay² + bx + cy + 1 = 0</p><p>Comparing with Ax² + 2Hxy + By² + 2Gx + 2Fy + C = 0:</p><p>A = 4, 2H = -4 → H = -2, B = a, 2G = b → G = b/2, 2F = c → F = c/2, C = 1</p><p><strong>Step 2:</strong> Apply the condition for identical straight lines: Δ = ABC + 2FGH - AF² - BG² - CH² = 0</p><p>Substituting values:</p><p>4·a·1 + 2·(c/2)·(b/2)·(-2) - 4·(c/2)² - a·(b/2)² - 1·(-2)² = 0</p><p>4a + (-bc/2) - c² - ab²/4 - 4 = 0</p><p>4a - bc/2 - c² - ab²/4 - 4 = 0</p><p><strong>Step 3:</strong> Also, for a pair of identical lines, the quadratic form 4x² - 4xy + ay² must be a perfect square.</p><p>For 4x² - 4xy + ay² = (px + qy)², we need: 4x² - 4xy + ay² = p²x² + 2pqxy + q²y²</p><p>Comparing: p² = 4 → p = ±2, 2pq = -4, q²y² = ay²</p><p>If p = 2: 2q = -2 → q = -1, so a = 1</p><p>If p = -2: -2q = -2 → q = 1, so a = 1</p><p>Therefore: a = 1</p><p><strong>Step 4:</strong> With a = 1, the equation becomes: 4x² - 4xy + y² + bx + cy + 1 = 0</p><p>This factors as: (2x - y)² + bx + cy + 1 = 0</p><p>Let u = 2x - y. Then: u² + bx + cy + 1 = 0</p><p>For identical lines (2x - y + α)² = 0 for some α:</p><p>4x² - 4xy + y² + 4αx - 2αy + α² = 0</p><p>Comparing: 4α = b, -2α = c, α² = 1</p><p>So α = ±1</p><p><strong>Step 5:</strong> Case 1: α = 1 gives b = 4, c = -2</p><p>Ordered triplet: (a₁, b₁, c₁) = (1, 4, -2)</p><p><strong>Step 6:</strong> Case 2: α = -1 gives b = -4, c = 2</p><p>Ordered triplet: (a₂, b₂, c₂) = (1, -4, 2)</p><p><strong>Step 7:</strong> Sum = a₁ + b₁ + c₁ + a₂ + b₂ + c₂ = 1 + 4 + (-2) + 1 + (-4) + 2 = 2</p><p><strong>∴ Answer: 2</strong></p>