If $\begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix} = k \cdot a^l b^m c^n$, then

Match List-I with List-II

List-I	List-II
(A) $l = m = n =$	(I) 10
(B) $k + l + m + n =$	(II) 6
(C) $k^2 + l^2 + (m - n)^2 =$	(III) 2
(D) $l^2 + m^2 + (n - k) =$	(IV) 20

Choose the correct answer from the options given below:

The determinant is:

$\begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix}$

Factor out $a$ from Row 1, $b$ from Row 2, and $c$ from Row 3:

$= abc \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix}$

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Expanding along Row 1:

$\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = -a \begin{vmatrix} -b & c \\ b & -c \end{vmatrix} - b \begin{vmatrix} a & c \\ a & -c \end{vmatrix} + c \begin{vmatrix} a & -b \\ a & b \end{vmatrix}$

Calculate each $2 \times 2$ determinant:

$\begin{vmatrix} -b & c \\ b & -c \end{vmatrix} = (-b)(-c) - (c)(b) = bc - bc = 0$

$\begin{vmatrix} a & c \\ a & -c \end{vmatrix} = a(-c) - (c)(a) = -ac - ac = -2ac$

$\begin{vmatrix} a & -b \\ a & b \end{vmatrix} = a(b) - (-b)(a) = ab + ab = 2ab$

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Substituting back:

$= -a(0) - b(-2ac) + c(2ab)$

$= 0 + 2abc + 2abc$

$= 4abc$

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The complete determinant value:

$\text{Determinant} = abc \times 4abc$

$= 4a^2b^2c^2$

Therefore: $k = 4$, $l = 2$, $m = 2$, $n = 2$

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(A) $l = m = n = 2$ matches with (III)

(B) $k + l + m + n = 4 + 2 + 2 + 2 = 10$ matches with (I)

(C) $k^2 + l^2 + (m - n)^2 = 4^2 + 2^2 + (2-2)^2 = 16 + 4 + 0 = 20$ matches with (IV)

(D) $l^2 + m^2 + (n - k) = 2^2 + 2^2 + (2-4) = 4 + 4 + (-2) = 6$ matches with (II)

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The correct matching is: (A) - (III), (B) - (I), (C) - (IV), (D) - (II)