The correct answer is A. Absolute value remains unchanged but sign will change.
The determinant of a 3×3 matrix can be computed using the following formula:
$$\det(A) = \sum_{i=1}^3 (-1)^{i+1} a_{1i} \det(A_{i,i})$$
where $A_{i,i}$ is the determinant of the matrix that is obtained from $A$ by deleting row $i$ and column $i$.
If any two columns of $A$ are interchanged, then the sign of the determinant will change. This is because the determinant of $A_{i,i}$ will be equal to the determinant of $A_{i’,i’}$, where $i’$ is the index of the column that was interchanged with column $i$. However, the order of the terms in the sum in the formula for the determinant will be different, so the sign of the determinant will change.
The absolute value of the determinant will not change, because the determinant of a matrix is always a real number.
Here is an example to illustrate this. Let $A$ be the following matrix:
$$A = \left| {\begin{array}{*{20}{c}} 1&2&3 \ 4&5&6 \ 7&8&9 \end{array}} \right|$$
If we interchange columns 1 and 2, then the new matrix is:
$$B = \left| {\begin{array}{*{20}{c}} 4&2&3 \ 1&5&6 \ 7&8&9 \end{array}} \right|$$
The determinant of $A$ is equal to $-30$, and the determinant of $B$ is also equal to $-30$. However, the sign of the determinant of $B$ is negative, because the order of the terms in the sum in the formula for the determinant is different.