Applying \(R_1 \to R_1 + R_3 - 2R_2\), we get \(|\begin{smallmatrix} 0 & 0 & a+c-2b \\ 3y+5 & 6y+8 & 9y+b \\ 4y+6 & 7y+9 & 10y+c \end{smallmatrix}|\) Since a, b, c are in A.P., \(2b = a+c\), therefore \(a+c-2b=0\). Hence, the value of the determinant is 0.