Let $f\left(x\right)=(x-a)(x-b)(x-c),a<b<c$. The number of roots $f^{\text{'}}\left(x\right)=0$ has belonging to $(a,b)$ and $(b,c)$ is

 Here, $f\left(x\right)$ being a polynomial, is continuous and differentiable for all real values of $x$.

It also have, $f\left(a\right)=f\left(b\right)=f\left(c\right)$

If Rolle's theorem is applied to $f\left(x\right)$ in $[a,b]$ and $[b,c]$ 

It is observed that $f^{\text{'}}\left(x\right)=0$ would have at least one root in $(a,b)$ and at least one root in $(b,c)$.

But $f^{\text{'}}\left(x\right)=0$ is a polynomial of degree two, hence $f^{\text{'}}\left(x\right)=0$ cannot have more than two roots.

Hence, the number of roots that $f^{\text{'}}\left(x\right)=0$ has is two roots.

Therefore, the correct answer is option $\left(3\right)$.