The value of beginvmatrix1&2&3\\2&3&4\\3&4&5endvmatrix is

Apply $R_2\to R_2-R_1$ and $R_3\to R_3-R_2$ (using the original $R_2$): both new rows become $(1,1,1)$, so two rows are identical and the determinant is $0$. In general, a determinant whose entries are in AP along rows and columns vanishes. Evaluating by cofactors without noticing this wastes time and invites sign errors.