Solve the following equations

Let $f(x)=\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|$.

Now,

$f(a)=\left|\begin{array}{ccc}0 & a-a & a-b \\ a+a & 0 & a-c \\ a+b & a+c & 0\end{array}\right|$

$=\left|\begin{array}{ccc}0 & 0 & a-b \\ 2 a & 0 & a-c \\ a+b & a+c & 0\end{array}\right|$

$=(a-b)\left(2 a^{2}+2 a c\right) \neq 0$

$f(b)=\left|\begin{array}{ccc}0 & b-a & b-b \\ b+a & 0 & b-c \\ b+b & b+c & 0\end{array}\right|$

$=\left|\begin{array}{ccc}0 & b-a & 0 \\ b+a & 0 & b-c \\ 2 a & b+c & 0\end{array}\right|$

$=(b-a)(2 a b-2 a c) \neq 0$

$f(0)=\left|\begin{array}{ccc}0 & 0-a & 0-b \\ 0+a & 0 & 0-c \\ 0+b & 0+c & 0\end{array}\right|$

$=\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|$

$=a(b c)-b(a c)=0$

Hence, the correct option is (c).