If $f(x) = \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix}$, then

$f(0) = 0$

The correct answer is Option (3) → $f(0) = 0$ ##

We have,

$f(x) = \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix} \quad \dots(i)$

On substituting $x = a$ in Eq. (i) we get

$\Rightarrow f(a) = \begin{vmatrix} 0 & 0 & a-b \\ 2a & 0 & a-c \\ a+b & a+c & 0 \end{vmatrix}$

On expanding along $R_1$, we get

$= [(a-b)(2a \cdot (a+c))] \neq 0$

On substituting $x = b$ in Eq. (i) we get

$∴f(b) = \begin{vmatrix} 0 & b-a & 0 \\ b+a & 0 & b-c \\ 2b & b+c & 0 \end{vmatrix}$

On expanding along $R_1$, we get

$= -(b-a)[-2b(b-c)]$

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

On substituting $x = 0$ in Eq. (i) we get

$∴f(0) = \begin{vmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{vmatrix}$

On expanding along $R_1$, we get

$= a(bc) - b(ac)$

$= abc - abc = 0$