If $$\begin{vmatrix}-2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c\end{vmatrix} = \alpha(a+b)(b+c)(c+a) \ne 0$$ then $$\alpha$$ is equal to

Let

$$D=\begin{vmatrix}-2a & a+b & a+c \\ a+b & -2b & b+c \\ a+c & b+c & -2c\end{vmatrix}.$$

Since

$$D=\alpha(a+b)(b+c)(c+a),$$

the determinant is a homogeneous polynomial of degree $$3$$ in $$a,b,c$$.

To determine the constant $$\alpha$$, it is sufficient to substitute convenient values of $$a,b,c$$ such that

$$(a+b)(b+c)(c+a)\neq0.$$

Choose

$$a=b=c=1.$$

Then

$$D=\begin{vmatrix}-2 & 2 & 2 \\ 2 & -2 & 2 \\ 2 & 2 & -2\end{vmatrix}.$$

For a matrix of order $$3$$ having diagonal entry $$x$$ and all off-diagonal entries equal to $$y,$$ the determinant is

$$(x-y)^2(x+2y).$$

Here

$$x=-2,\qquad y=2.$$

Therefore,

$$D=(-2-2)^2(-2+2\cdot2).$$

Hence,

$$D=(-4)^2(2)=32.$$

Also,

$$(a+b)(b+c)(c+a)=(1+1)(1+1)(1+1)=8.$$

Using

$$D=\alpha(a+b)(b+c)(c+a),$$

we get

$$32=8\alpha.$$

Therefore,

$$\alpha=4.$$