(C) Given the determinant: $\Delta = \begin{vmatrix} 2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1 \end{vmatrix}$.Expanding along the first row:$\Delta = 2

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(C) Given the determinant: $\Delta = \begin{vmatrix} 2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1 \end{vmatrix}$.Expanding along the first row:$\Delta = 2[(k-1)(k+1) - 1] - 2k[1(k+1) - 2] + 1[1 - 2(k-1)]$$\Delta = 2[k^2 - 1 - 1] - 2k[k + 1 - 2] + 1[1 - 2k + 2]$$\Delta = 2[k^2 - 2] - 2k[k - 1] + [3 - 2k]$$\Delta = 2k^2 - 4 - 2k^2 + 2k + 3 - 2k$$\Delta = -1$.Comparing $\Delta = -1$ with $Ak^2 + Bk + C$, we get $A = 0, B = 0, C = -1$.Therefore, $A + B + C = 0 + 0 - 1 = -1$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\left|\begin{array}{ccc}2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1\end{array}\right|=Ak^2+Bk+C$, then $A+B+C=$

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$0$
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$1$
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$-1$
D
$2$

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