Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} ; \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$ then we get
$\Delta=(a+b+c)^{2}\left|\begin{array}{ccc}{(b+c)^{2}} & {a-(b+c)} & {a-(b+c)} \\ {b^{2}} & {(a+c)-b} & {0} \\ {c^{2}} & {0} & {(a+b)-c}\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\left(\mathrm{R}_{2}+\mathrm{R}_{3}\right)$
$\Delta=(a+b+c)^{2}\left|\begin{array}{ccc}{2 b c} & {-2 c} & {-2 b} \\ {b^{2}} & {a+c-b} & {0} \\ {c^{2}} & {0} & {a+b-c}\end{array}\right|$
Now applying $\mathrm{C}_{2} \rightarrow\left(\mathrm{C}_{2}+\frac{1}{\mathrm{b}} \mathrm{C}_{1}\right)$ and
${{{\rm{C}}_3} \to \left( {{{\rm{C}}_3} + \frac{1}{{\rm{c}}}{{\rm{C}}_1}} \right);{\rm{ we get }}}$
$\Delta = {\left( {a + b + c} \right)^2}\left| {\begin{array}{*{20}{c}}
{2bc}&0&0\\
{{b^2}}&{a + c}&{\frac{{{b^2}}}{c}}\\
{{c^2}}&{\frac{{{c^2}}}{b}}&{a + b}
\end{array}} \right|$
${ \Rightarrow {\rm{ So }}\Delta = 2abc{{(a + b + c)}^3}}$
Method-2 $\quad$ Put $a=1, b=1$ and $c=1$ and check from option
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