The value of the determinant \[\left| {\begin{array}{*{20}{c}} 1&3&2 \\ 4&1&1 \\ 2&1&3 \end{array}} \right|\] is A. -28 B. -24 C. -32 D. 36

The correct answer is $\boxed{-28}$.

To find the determinant of a 3×3 matrix, we can use the formula below:

$$\left| {\begin{array}{*{20}{c}} a&b&c \ d&e&f \ g&h&i \end{array}} \right| = (aei – bghf – chf + dgi) + (bfi – ceg – ahf + dhi) + (cdi – aeg – bhf + bg)$$

In this case, we have:

$$\left| {\begin{array}{*{20}{c}} 1&3&2 \ 4&1&1 \ 2&1&3 \end{array}} \right| = (1 \cdot 3 \cdot 3 – 3 \cdot 1 \cdot 2 – 2 \cdot 4 \cdot 1) + (3 \cdot 1 \cdot 3 – 1 \cdot 4 \cdot 2 – 2 \cdot 2 \cdot 1) + (2 \cdot 1 \cdot 3 – 1 \cdot 1 \cdot 2 – 3 \cdot 4)$$

$$= (9 – 6 – 8) + (3 – 8 – 4) + (6 – 2 – 12)$$

$$= -28$$