$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{bc}&{a(b + c)}\\1&{ca}&{b(c + a)}\\1&{ab}&{c(a + b)}\end{array}} \right| = 0$

L.H.S. $=$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{bc}&{a(b + c)}\\1&{ca}&{b(c + a)}\\1&{ab}&{c(a + b)}\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{bc}&{ab + ac}\\1&{ca}&{bc + ba}\\1&{ab}&{ca + cb}\end{array}} \right| = 0$

Applying ${C_3} \to {C_3} + {C_2},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{bc}&{ab + bc + ac}\\1&{ca}&{bc + ca + ab}\\1&{ab}&{ca + cb + ab}\end{array}} \right| = (ab + bc + ca)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{bc}&1\\1&{ca}&1\\1&{ab}&1\end{array}} \right|$

$= (ab + bc + ca) \times 0 = 0$ [as ${C_1} \sim {C_3}$]