Evaluate :

$\Delta=\left|\begin{array}{lll}a & b+c & a^{2} \\ b & c+a & b^{2} \\ c & a+b & c^{2}\end{array}\right|$

When $a=b$, the first two rows become identical. Hence, $a-b$ is a factor.

Similarly, when $b=c$ the second and third rows become identical. So, $b-c$ is also a factor.

Also, when $c=a$, the third and first rows become identical. Hence, $c-a$ is also a factor.

The product of diagonal elements, $\mathrm{a}(\mathrm{c}+\mathrm{a}) \mathrm{c}^{2}$ is 4 . So, the other factor should be a linear in $a, b$ and $c .$ It should also remain

unaltered when any two letters are changed. Let this factor be $\lambda(\mathrm{a}+\mathrm{b}+\mathrm{c})$.

Here, $\lambda$ is a constant. To find this, we have

$a=0, b=1, c=2$

$\left|\begin{array}{lll}0 & 3 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 4\end{array}\right|=\lambda(a-b)(b-c)(c-a)(a+b+c)$

$\left|\begin{array}{lll}0 & 3 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 4\end{array}\right|=\lambda(0-1)(1-2)(2-1)(0+1+2)$

$\Rightarrow-6=6 \lambda$

$\Rightarrow \lambda=-1$

Thus,

$\left|\begin{array}{lll}a & b+c & a^{2} \\ b & c+a & b^{2} \\ c & a+b & c^{2}\end{array}\right|=-\left(\left(\begin{array}{l}a+b+c \\ a+b\end{array}\right)\right)\left(\begin{array}{l}a-c \\ a-b-a\end{array}\right)\left(\begin{array}{c}c-a \\ c-a\end{array}\right)$