Multiply ${C_1}$ by $a,{C_2}$by $b$ and ${C_3}$ by $c$ and hence divide by $abc$.
$ = \frac{1}{{abc}}\left| {\begin{array}{*{20}{c}}
{a\left( {{b^2} + {c^2}} \right)}&{a{b^2}}&{a{c^2}}\\
{{a^2}b}&{b\left( {{c^2} + {a^2}} \right)}&{b{c^2}}\\
{{a^2}c}&{{b^2}c}&{c\left( {{a^2} + {b^2}} \right)}
\end{array}} \right|$
Take out $a,b,c $ common fopre ${R_1},{R_2}$ and ${R_3}$ respectively.
$\Delta = \frac{{abc}}{{abc}}\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{{b^2}}&{{c^2}}\\
{{a^2}}&{{c^2} + {a^2}}&{{c^2}}\\
{{a^2}}&{{b^2}}&{{a^2} + {b^2}}
\end{array}} \right|$
Apply ${C_1} \to {C_1} - {C_2} - {C_3}$
$\Delta = \left| {\begin{array}{*{20}{c}}
0&{{b^2}}&{{c^2}}\\
{ - 2{c^2}}&{{c^2} + {a^2}}&{{c^2}}\\
{ - 2{b^2}}&{{b_2}}&{{a^2} + {b^2}}
\end{array}} \right|$
$ = - 2\left| {\begin{array}{*{20}{c}}
0&{{b^2}}&{{c^2}}\\
{{c^2}}&{{c^2} + {a^2}}&{{c^2}}\\
{{b^2}}&{{b^2}}&{{a^2} + {b^2}}
\end{array}} \right|$
Apply ${C_2} - {C_1}$ and ${C_3} - {C_1}$
$ = - 2\left| {\begin{array}{*{20}{c}}
0&{{b^2}}&{{c^2}}\\
{{c^2}}&{{a^2}}&0\\
{{b^2}}&0&{{a^2}}
\end{array}} \right|$
$ = - 2\left[ { - {b^2}\left( {{c^2}{a^2}} \right) + {c^2}\left( { - {a^2}{b^2}} \right)} \right]$
$ = 2{a^2}{b^2}{c^2} + 2{a^2}{b^2}{c^2} = 4{a^2}{b^2}{c^2}$
But $\Delta = k{a^2}{b^2}{c^2}\therefore k = 4$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.