If a, b, c are the roots of equation <span class="math-tex">\(x^3-3x^2 + 3x + 7=0\)</span>, then the value of <span class="math-tex">\(\begin{vmatrix} 2bc-a^2 & c^2 & b^2 \\\ c^2 & 2ac-b^2 & a^2 \\\ b^2 & a^2 & 2ab-c^2 \end{vmatrix}\)</span> is
This determinant equals the square of \(\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}\), so it is \((a^3 + b^3 + c^3 - 3abc)^2\).
From the equation: \(a + b + c = 3\), \(ab + bc + ca = 3\), \(abc = -7\).
\(a^2 + b^2 + c^2 = 9 - 6 = 3\)
\(a^3 + b^3 + c^3 - 3abc = (a+b+c)\left[(a^2+b^2+c^2) - (ab+bc+ca)\right] = 3(3 - 3) = 0\)
So the determinant = 0.
Answer: (D) 0
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