If $Δ (x) = |\begin{array}{ccc} x^{2} - 5 x + 3 & 2 x - 5 & 3 \\ 3 x^{2} + x + 4 & 6 x + 1 & 9 \\ 7 x^{2} - 6 x + 9 & 14 x - 6 & 21 \end{array}|$ $= {ax}^{3} + {bx}^{2} + cx + d$ , then

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$a = 0$

$b = 0$

$c = 0$

$d = 47$

answer is A.

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Detailed Solution

$∵ Δ (x) = |\begin{array}{ccc} x^{2} - 5 x + 3 & 2 x - 5 & 3 \\ 3 x^{3} + x + 4 & 6 x + 1 & 9 \\ 7 x^{2} - 6 x + 9 & 14 x - 6 & 21 \end{array}| Applying R_{2} \to R_{2} - 3 R_{1} and R_{3} \to R_{3} - 7 R_{1}, then = |\begin{array}{ccccc} x^{2} - 5 x + 3 & \dots & 2 x - 5 & \dots & 3 \\ ⋮ \\ 16 x - 5 & 16 & 0 \\ ⋮ \\ 29 x - 12 & 29 & 0 \end{array}| Expanding along C_{3}, we get = 3 |\begin{array}{cc} 16 x - 5 & 16 \\ 29 x - 12 & 29 \end{array}| Applying C_{1} \to C_{1} - {xC}_{2}, then Δ (x) = 3 |\begin{array}{cc} - 5 & 16 \\ - 12 & 29 \end{array}| = 3 (- 145 + 192) = 3 \times 47 = 141 = {ax}^{3} + {bx}^{2} + cx + d [given] ∴ a = 0, b = 0, c = 0, d = 141$