Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant \[\left| \left. \begin{matrix}
   -2a & a+b & a+c \\
   b+a & -2b & b+c \\
   c+a & c+b & -2c \\
\end{matrix} \right| \right.\] . The other factor in the value of the determinant

Hint: To solve this problem first we have to make the necessary assumption to simplify the given terms and then we have to apply some of the row and column operations to solve the determinant. Then the simplified determinant will leave us with the required answer.

Complete step-by-step solution:
Let \[\Delta =\left| \left. \begin{matrix}
   -2a & a+b & a+c \\
   b+a & -2b & b+c \\
   c+a & c+b & -2c \\
\end{matrix} \right| \right.\]
And let $a + b = 2C, b + c = 2A, c + a = 2B$
\[\Rightarrow \] $a + b + b + c + c + a = 2A + 2B + 2C$
\[\Rightarrow \] \[2\left( a+b+c \right)\] = \[2\left( A+B+C \right)\]
\[\Rightarrow \] \[\left( a+b+c \right)\] = \[\left( A+B+C \right)\]
Also we have,
\[a=\left( a+b+c \right)-\left( b+c \right)\] = \[\left( A+B+C \right)-2A\] = \[B+C-A\]
Similarly we get,
\[b=C+A-B\]
\[c=A+B-C\]
Now on substitution in \[\Delta \] , we get
\[\Delta =\left| \left. \begin{matrix}
   2A-2B-2C & 2C & 2B \\
   2C & 2B-2C-2A & 2A \\
   2B & 2A & 2C-2A-2B \\
\end{matrix} \right| \right.\]
We can take 2 common from \[{{R}_{1}},{{R}_{2}},{{R}_{3}}\]
\[\Rightarrow 8\times \left| \left. \begin{matrix}
   A-B-C & C & B \\
   C & B-C-A & A \\
   B & A & C-A-B \\
\end{matrix} \right| \right.\]
On applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}},{{C}_{2}}\to {{C}_{2}}+{{C}_{3}}\] , we get
\[\Rightarrow 8\times \left| \left. \begin{matrix}
   A-B & C+B & B \\
   B-A & B-C & A \\
   B+A & C-B & C-A-B \\
\end{matrix} \right| \right.\]
On applying \[{{R}_{2}}\to {{R}_{2}}+{{R}_{1}},{{R}_{3}}\to {{R}_{3}}+{{R}_{2}}\] , we get
\[\Rightarrow 8\times \left| \left. \begin{matrix}
   A-B & B+C & B \\
   0 & 2B & A+B \\
   2B & 0 & C-B \\
\end{matrix} \right| \right.\]
Now let us expand along \[{{C}_{1}}\]
\[\Rightarrow 8\times \left\{ \left( A-B \right)\left| \left. \begin{matrix}
   2B & A+B \\
   0 & C-B \\
\end{matrix} \right|+\left( 2B \right)\left| \left. \begin{matrix}
   C+B & B \\
   2B & A+B \\
\end{matrix} \right| \right. \right. \right\}\]
\[\Rightarrow 16B\left\{ \left( A-B \right)\left( C-B \right)+\left( C+B \right)\left( A+B \right)-2{{B}^{2}} \right\}\]
Now let us simplify this to get as below,
\[\Rightarrow 32ABC\]
We should re substitute the values of A, B and C,
\[\Rightarrow 32\left( \dfrac{b+c}{2} \right)\left( \dfrac{c+a}{2} \right)\left( \dfrac{a+b}{2} \right)\]
\[\Rightarrow 4\left( b+c \right)\left( c+a \right)\left( a+b \right)\]
Hence 4 is the other factor of the determinant.

Note: Students should be aware of the row and column operation to be applied to get the simplified form of the determinant. We can also solve this determinate only by basic expansion of the determinant but sometimes students may get confused and miss the terms while expanding.