A mixed determinant exercise

Add columns. $C_1\leftarrow C_1+C_2+C_3$: each entry in the first column becomes $2(a+b+c)$. So
$$\Delta = 2(a+b+c)\begin{vmatrix} 1 & b+c & c+a \\ 1 & c+a & a+b \\ 1 & a+b & b+c \end{vmatrix}.$$

Row operations. $L_2\leftarrow L_2-L_1$, $L_3\leftarrow L_3-L_1$:
$$\Delta = 2(a+b+c)\begin{vmatrix} 1 & b+c & c+a \\ 0 & a-b & b-c \\ 0 & a-c & b-a \end{vmatrix}.$$

Expand along column 1.
$$\Delta = 2(a+b+c)\begin{vmatrix} a-b & b-c \\ a-c & b-a \end{vmatrix}.$$

$2\times2$ determinant.
$$(a-b)(b-a) - (b-c)(a-c) = -(a-b)^2 - (b-c)(a-c).$$
Expand:
$$= -(a^2-2ab+b^2) - (ab-ac-b^2+bc) = -a^2+2ab-b^2-ab+ac+b^2-bc = -a^2+ab+ac-bc = -(a-b)(a-c).$$

Conclusion.
$$\Delta = -2(a+b+c)(a-b)(a-c).$$

Symmetry check. The result should be symmetric in $a,b,c$. Since $(a-b)(a-c)$ is not symmetric, but $(a+b+c)(a-b)(a-c)$ is antisymmetric under swapping $b,c$, the expression is consistent with the alternating property of the determinant.