Let a, b, c be positive real numbers.

$(\mathrm{b})$ unique solution

The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{ccc}\frac{1}{a^{2}} & \frac{1}{b^{2}} & \frac{-1}{c^{2}} \\ \frac{1}{a^{2}} & \frac{-1}{b^{2}} & \frac{1}{c^{2}} \\ \frac{-1}{a^{2}} & \frac{1}{b^{2}} & \frac{1}{c^{2}}\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

Here,

$A=\left[\begin{array}{ccc}\frac{1}{a^{2}} & \frac{1}{b^{2}} & \frac{-1}{c^{2}} \\ \frac{1}{a^{2}} & \frac{-1}{b^{2}} & \frac{1}{c^{2}} \\ \frac{-1}{a^{2}} & \frac{1}{b^{2}} & \frac{1}{c^{2}}\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ccc}\frac{1}{a^{2}} & \frac{1}{b^{2}} & \frac{-1}{c^{2}} \\ \frac{1}{a^{2}} & \frac{-1}{b^{2}} & \frac{1}{c^{2}} \\ \frac{-1}{a^{2}} & \frac{1}{b^{2}} & \frac{1}{c^{2}}\end{array}\right|$

$=\frac{1}{a^{2} b^{2} c^{2}}\left|\begin{array}{ccc}1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1\end{array}\right|$

$=\frac{1}{a^{2} b^{2} c^{2}} \times 1(-1-1)-1(1+1)-1(1-1)$

$=\frac{1}{a^{2} b^{2} c^{2}} \times(-2-2)$

$=\frac{-4}{a^{2} b^{2} c^{2}}$

$\Rightarrow|A| \neq 0$

So, the given system of equations has a unique solution.