<p>If \(\Delta = \begin{vmatrix} 3 & 4 & 5 & x \\ 4 & 5 & 6 & y \\ 5 & 6 & 7 & z \\ x & y & z & 0 \end{vmatrix} = 0\), then</p>
<p>\(x, y, z\) are in A.P.</p>
<p>\(x, y, z\) are in G.P.</p>
<p>\(x, y, z\) are in H.P.</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that the first three rows form an arithmetic progression pattern in each column, making them linearly dependent. Use row operations to expose this dependency and establish the constraint on x, y, z.
<p><strong>Step 1:</strong> Observe the pattern in rows 1-3: Row 2 - Row 1 = (1,1,1,y-x) and Row 3 - Row 2 = (1,1,1,z-y).</p><p><strong>Step 2:</strong> Perform R₂ → R₂ - R₁ and R₃ → R₃ - R₂:</p><p>The matrix becomes:</p><p>$\begin{vmatrix} 3 & 4 & 5 & x \\ 1 & 1 & 1 & y-x \\ 1 & 1 & 1 & z-y \\ x & y & z & 0 \end{vmatrix}$</p><p><strong>Step 3:</strong> Notice rows 2 and 3 become identical when y-x = z-y, i.e., 2y = x+z (or x, y, z are in A.P.)</p><p><strong>Step 4:</strong> If the determinant equals zero and rows 2,3 have the pattern shown, the constraint is: <strong>x + z = 2y</strong> (x, y, z are in arithmetic progression)</p><p>∴ Answer: A</p>
Correct Answer: A