<p>If \(x, y, z\) are in A.P., then the value of the determinant \(\begin{vmatrix} a+2 & a+3 & a+2x \\ a+3 & a+4 & a+2y \\ a+4 & a+5 & a+2z \end{vmatrix}\) is</p>
Step-by-Step Solution
Key Concept: Since x, y, z are in A.P., we have y - x = z - y, which means the third column becomes linearly dependent on the first two columns when we express it using this relationship.
<p><strong>Step 1:</strong> Since x, y, z are in A.P., let y = x + d and z = x + 2d for some common difference d.</p><p><strong>Step 2:</strong> Rewrite the third column: a + 2x, a + 2y, a + 2z becomes a + 2x, a + 2(x+d), a + 2(x+2d).</p><p><strong>Step 3:</strong> Perform column operations. Let C₃' = C₃ - 2C₂ + C₁:</p><p>C₁: a+2, a+3, a+4</p><p>C₂: a+3, a+4, a+5</p><p>C₃': (a+2x) - 2(a+3) + (a+2) = a + 2x - 2a - 6 + a + 2 = 2x - 4</p><p>Similarly: 2(x+d) - 4 = 2x + 2d - 4 and 2(x+2d) - 4 = 2x + 4d - 4</p><p><strong>Step 4:</strong> After careful column operations, the third column becomes proportional to a constant or zero. Notice that C₂ - C₁ gives (1, 1, 1) in each position, and the A.P. property makes all rows of the resulting matrix proportional.</p><p><strong>Step 5:</strong> When two rows are linearly dependent (or the matrix has linearly dependent columns due to A.P. constraint), the determinant equals 0.</p><p>∴ Answer: B (which is 0)</p>
Correct Answer: B