Answer :

To find the determinant of the given [tex]\(3 \times 3\)[/tex] matrix:
[tex]\[ \begin{vmatrix} 1 & 3 & 1 \\ 1 & -2 & -1 \\ 2 & 1 & 2 \end{vmatrix} \][/tex]
we can use the rule for determinants of [tex]\(3 \times 3\)[/tex] matrices, which is given by:

[tex]\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \][/tex]

In our matrix:

[tex]\[ \begin{array}{ccc} a = 1 & b = 3 & c = 1 \\ d = 1 & e = -2 & f = -1 \\ g = 2 & h = 1 & i = 2 \end{array} \][/tex]

Substitute these values into the determinant formula:

[tex]\[ \begin{vmatrix} 1 & 3 & 1 \\ 1 & -2 & -1 \\ 2 & 1 & 2 \end{vmatrix} = 1((-2 \cdot 2) - (-1 \cdot 1)) - 3((1 \cdot 2) - (-1 \cdot 2)) + 1((1 \cdot 1) - (-2 \cdot 2)) \][/tex]

Now, calculate each term within the parentheses:

[tex]\[ 1((-4) - (-1)) - 3((2) - (-2)) + 1((1) - (-4)) \][/tex]

Simplify the expressions in the parentheses:

[tex]\[ 1(-4 + 1) - 3(2 + 2) + 1(1 + 4) \][/tex]

[tex]\[ 1(-3) - 3(4) + 1(5) \][/tex]

Then, perform the multiplications:

[tex]\[ -3 - 12 + 5 \][/tex]

Finally, calculate the sum of these terms:

[tex]\[ -3 - 12 + 5 = -10 \][/tex]

Thus, the determinant of the matrix is:

[tex]\[ \boxed{-10} \][/tex]