Let $\Delta  = \left| {\begin{array}{*{20}{c}}
{ – 2a}&{a + b}&{a + c}\\
{b + a}&{ – 2b}&{b + c}\\
{c + a}&{b + c}&{ – 2c}
\end{array}} \right|$

Applying ${C_1} + {C_3}$ and ${C_2} + {C_3}$

$\Delta  = \left| {\begin{array}{*{20}{c}}
{ – a + c}&{2a + b + c}&{a + c}\\
{2b + a + c}&{ – b + c}&{b + c}\\
{a – c}&{b – c}&{ – 2c}
\end{array}} \right|$

Now, applying ${R_1} + {R_3}$ and ${R_2} + {R_3}$

$\Delta  = \left| {\begin{array}{*{20}{c}}
0&{2\left( {a + b} \right)}&{a – c}\\
{2\left( {a + b} \right)}&0&{b – c}\\
{a – c}&{b – c}&{ – 2c}
\end{array}} \right|$

On expanding, we get

$\Delta  =  – 2\left( {a + b} \right)\left\{ { – 2c\left[ {2\left( {a + b} \right)} \right] – \left( {a – c} \right)\left( {b – c} \right)} \right\}$ $ + \left( {a – c} \right)\left[ {2\left( {a + b} \right)\left( {b – c} \right)} \right]$

$\Delta  = 8c\left( {a + b} \right)\left( {a + b} \right) + 4\left( {a + b} \right)\left( {a – c} \right)\left( {b – c} \right)$

$ = 4\left( {a + b} \right)\left[ {2ac + 2bc + ab – bc – ac + {c^2}} \right]$

$ = 4\left( {a + b} \right)\left[ {ac + bc + ab + {c^2}} \right]$

$ = 4\left( {a + b} \right)\left[ {c\left( {a + c} \right) + b\left( {a + c} \right)} \right]$

$ = 4\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$

$ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$

Hence, $\alpha  = 4$