If $f(x) = \begin{vmatrix} 0 & x-1 & x-2 \\ x+1 & 0 & x-3 \\ x+2 & x+3 & 0 \end{vmatrix}$, then the value of $f(0)$ is equal to:

$f(x) = \begin{vmatrix} 0 & x-1 & x-2 \\ x+1 & 0 & x-3 \\ x+2 & x+3 & 0 \end{vmatrix}$

To find $f(0)$, substitute $x = 0$ into the matrix:

$f(0) = \begin{vmatrix} 0 & 0-1 & 0-2 \\ 0+1 & 0 & 0-3 \\ 0+2 & 0+3 & 0 \end{vmatrix}$

$f(0) = \begin{vmatrix} 0 & -1 & -2 \\ 1 & 0 & -3 \\ 2 & 3 & 0 \end{vmatrix}$

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Expanding along the first row:

$f(0) = (0) \cdot \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix}$

$f(0) = 0 + 1 \cdot \begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix}$

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For the first $2 \times 2$ determinant:

$\begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} = (1)(0) - (-3)(2)$

$= 0 + 6$

$= 6$

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For the second $2 \times 2$ determinant:

$\begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix} = (1)(3) - (0)(2)$

$= 3 - 0$

$= 3$

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$f(0) = 1(6) - 2(3)$

$f(0) = 6 - 6$

$f(0) = 0$