<p>If \(a, b\) and \(c\) are sides of a triangle and \(\begin{vmatrix} a^2 & b^2 & c^2 \\ (a+1)^2 & (b+1)^2 & (c+1)^2 \\ (a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix} = 0\), then</p>
<p>(a) \(\triangle ABC\) is an equilateral triangle</p>
<p>(b) \(\triangle ABC\) is a right angled isosceles triangle</p>
<p>(c) \(\triangle ABC\) is an isosceles triangle</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: When a determinant equals zero, the rows (or columns) are linearly dependent. We can use row operations to simplify the determinant and find relationships between a, b, and c.
<p><strong>Step 1: Simplify using row operations.</strong></p><p>Let the rows be R₁, R₂, R₃. Compute R₂ - R₁:</p><p>R₂ - R₁ = [(a+1)² - a², (b+1)² - b², (c+1)² - c²]</p><p>= [2a + 1, 2b + 1, 2c + 1]</p><p><strong>Step 2: Compute R₃ - R₁.</strong></p><p>R₃ - R₁ = [(a-1)² - a², (b-1)² - b², (c-1)² - c²]</p><p>= [-2a + 1, -2b + 1, -2c + 1]</p><p><strong>Step 3: Observe the new determinant.</strong></p><p>The determinant becomes:</p><p>∣ a² b² c² ∣</p><p>∣ 2a+1 2b+1 2c+1 ∣ = 0</p><p>∣ -2a+1 -2b+1 -2c+1 ∣</p><p><strong>Step 4: Further simplification.</strong></p><p>Notice that R₂ + R₃ = [1, 1, 1] (all entries equal to 1). If the determinant is zero and we have rows that become linearly dependent, consider: R₂ = 2(a, b, c) + (1, 1, 1) and R₃ = -(2(a, b, c) - (1, 1, 1)).</p><p>For the determinant to be zero with these constraints, the only possibility is when rows become proportional or when special relationships hold.</p><p><strong>Step 5: Apply the condition directly.</strong></p><p>Expanding along the structure: since det = 0, and examining the coefficients, we need:</p><p>(2a+1)(−2c+1) − (2c+1)(−2a+1) applied with first row elements must yield zero for all combinations.</p><p>This forces: a = b = c</p><p><strong>Step 6: Verify the conclusion.</strong></p><p>If a = b = c, the determinant becomes:</p><p>∣ a² a² a² ∣</p><p>∣ (a+1)² (a+1)² (a+1)² ∣ = 0 ✓</p><p>∣ (a-1)² (a-1)² (a-1)² ∣</p><p>All rows are proportional (columns identical), confirming det = 0.</p><p>When a = b = c for sides of a triangle, the triangle is equilateral.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A