Given, system of linear equation, $a b c \neq 0$, is
$$
x+7 a y+2 a z=0, x+6 b y+2 b z=0
$$
and $x+5 c y+2 c z=0$ has a non-zero trivial solution.
$$
\begin{aligned}
& \text { So, } A=0 \Rightarrow\left|\begin{array}{ccc}
1 & 7 a & 2 a \\
1 & 6 b & 2 b \\
1 & 5 c & 2 c
\end{array}\right|=0 \\
& \Rightarrow 1(12 b c-10 b c)-7 a(2 c-2 b)+2 a(5 c-6 b)=0 \\
& \Rightarrow 2 b c-4 a c+2 a b=0 \\
& \Rightarrow \quad 2 a c=b c+a b \\
& \Rightarrow \quad b=\frac{2 a c}{a+c} \text { or } \frac{2}{b}=\frac{1}{a}+\frac{1}{c}
\end{aligned}
$$

Then, $a, b, c$ are in Harmonic Progression.
Hence, option (a) is correct.