For the following system, $$\dot{x_1} = 3(x_1-x_2) \\ \dot{x_2} = x_1(k-2-x_3) \\ \dot{x_3}=x_1x_3$$ I want to determine its equilibrium points together with their stability.
By the third equation I get $x_1=0$ or $x_3 = 0$. If $x_1 =0$ then $x_2=0$ so that I found an equilibrium point $(0,0,0)$ as $x_1=x_2$ by the first equation. However, if $x_3=0$ then $x$ can be anything unless $(b-1-x_3)=0$ which means that $b=1$. So that if $b \neq 1$, I find equilibrium points $(x_1,x_2,0)=(x_1,x_1,0)$ where $x_1 \neq 0$ which looks unusual.
How can I find the equilibrium points of this system?
From $\dot{x}_3 = x_1x_3$ we have that either $x_1=0$ or $x_3=0$. Let us start with $x_1=0$. Then from $\dot{x}_1=3(x_1-x_2)$ we conclude $x_2=0$. For $x_1=0$ we have $\dot{x}_2=0$, and thus we obtain the equilibrium set $\left\{x: x_1=x_2=0, \ x_3\in \mathbb{R}\right\}$. Any point on this line is an equilibrium.
Next, consider $x_3=0$. From the equation for $\dot{x}_1$ we have $x_1=x_2$, and if $x_1=0$ we have the origin that belongs to the equilibrium set we have already found. Let us consider $x_1\ne 0$ to check if there is something else besides that line.
From the equation for $\dot{x}_2$ we conclude that the only solution satisfying $x_3=0$ and $x_1\ne 0$ is when $k=2$ that yields $x_1\in\mathbb{R}$.
To summarize, you have the equilibrium set (line) $\left\{x: x_1=x_2=0, \ x_3\in \mathbb{R}\right\}$. And for $k=2$ you have additionally the equilibrium set $\{x:x_1\in \mathbb{R}, x_2=x_1, x_3=0\}$ that is another line.