To solve the given problem, we first note that the question involves evaluating the determinant of a 3x3 matrix. We are given:

We are told that \(a\), \(b\), and \(c\) are in an arithmetic progression. In an arithmetic progression, the terms follow the relation:

\(b = \frac{a + c}{2}\)

This relationship will be crucial in simplifying the determinant. Now, let's denote the matrix by \(M\):

\[ M = \begin{vmatrix} x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix} \]

The determinant of a 3x3 matrix \(\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}\) is given by:

\(a(ei - fh) - b(di - fg) + c(dh - eg)\)

Applying this formula to our matrix \(M\), we simplify the calculation by focusing on the properties of determinants.

Since \(a\), \(b\), and \(c\) are in an arithmetic progression, we substitute \(b = \frac{a + c}{2}\). Evaluate the effect of this substitution:

After substituting for \(b\), notice each column elements being in an arithmetic progression:

\((x+2, x+3, x+4)\), \((x+3, x+4, x+5)\), and \((x+2a, x+2b, x+2c)\) simplifies to a linear combination.

Thus, the columns are linearly dependent, leading the determinant to be zero, as linearly dependent rows or columns in a matrix always yield a determinant of zero.

Hence, the value of the determinant is 0, showing that the correct answer is 0.