If a , b , c are the sides of a triangle ABC and p is the perimeter...

Let $\Delta = \begin{vmatrix} p+c \text{ and } a \text{ and } b \\ c \text{ and } p+a \text{ and } b \\ c \text{ and } a \text{ and } p+b \end{vmatrix}$

Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:

$\Delta = \begin{vmatrix} p+a+b+c \text{ and } a \text{ and } b \\ p+a+b+c \text{ and } p+a \text{ and } b \\ p+a+b+c \text{ and } a \text{ and } p+b \end{vmatrix}$

Since $p = a+b+c$ is the perimeter of the triangle, we have $p+a+b+c = 2p$. Substituting this into the determinant:

$\Delta = \begin{vmatrix} 2p \text{ and } a \text{ and } b \\ 2p \text{ and } p+a \text{ and } b \\ 2p \text{ and } a \text{ and } p+b \end{vmatrix}$

Taking $2p$ common from the first column $C_1$:

$\Delta = 2p \begin{vmatrix} 1 \text{ and } a \text{ and } b \\ 1 \text{ and } p+a \text{ and } b \\ 1 \text{ and } a \text{ and } p+b \end{vmatrix}$

Applying the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:

$\Delta = 2p \begin{vmatrix} 1 \text{ and } a \text{ and } b \\ 0 \text{ and } p \text{ and } 0 \\ 0 \text{ and } 0 \text{ and } p \end{vmatrix}$

Expanding the determinant along the first column $C_1$:

$\Delta = 2p [1(p \cdot p - 0)] = 2p(p^2) = 2p^3$

Answer: $2p^3$