<p>If <em>a, b, c, d, e,</em> and <em>f</em> are in G.P., then the value of \(\begin{vmatrix} a^2 & d^2 & x \\ b^2 & e^2 & y \\ c^2 & f^2 & z \end{vmatrix}\) depends on</p>
<p>(1) \(x\) and \(y\)</p>
<p>(2) \(x\) and \(z\)</p>
<p>(3) \(y\) and \(z\)</p>
<p>(4) independent of \(x\), \(y\) and \(z\)</p>
Step-by-Step Solution
Key Concept: Since a, b, c, d, e, f are in G.P., the second column is always a constant multiple of the first column (specifically, d²=a²r², e²=b²r², f²=c²r² where r is the common ratio). This makes columns 1 and 2 linearly dependent, collapsing the determinant structure regardless of x, y, z.
<p><strong>Step 1:</strong> Since a, b, c, d, e, f are in G.P., let the common ratio be r.</p><p>Then: d = ar, e = br, f = cr</p><p><strong>Step 2:</strong> Square these relations:</p><p>d² = a²r², e² = b²r², f² = c²r²</p><p><strong>Step 3:</strong> Observe that Column 2 = r² × Column 1:</p><p>$$\begin{vmatrix} a^2 & r^2a^2 & x \\ b^2 & r^2b^2 & y \\ c^2 & r^2c^2 & z \end{vmatrix}$$</p><p><strong>Step 4:</strong> Since C₂ = r²·C₁, columns 1 and 2 are linearly dependent.</p><p><strong>Step 5:</strong> By the property of determinants, if two columns are proportional, the determinant equals zero regardless of the third column.</p><p>$$\Delta = 0$$</p><p><strong>Step 6:</strong> The determinant is independent of x, y, and z. It depends only on the G.P. property, which forces linear dependence.</p><p>∴ Answer: <strong>D</strong> (Neither x, y, nor z individually; the determinant is always 0)</p>
Correct Answer: D