Problem. Let $f_i:\Bbb{R}^4\to \Bbb{R}, \,\, i=1,2,3,$ be defined by $$f_1(x_1,x_2,x_3,x_4) = x_1x_3-x_2^2\\f_2(x_1,x_2,x_3,x_4)=x_2x_4-x_3^2\\f_3(x_1,x_2,x_3,x_4)=x_1x_4-x_2x_3.$$ Then $M=\{x\in \Bbb{R}^4\setminus\{0\}: f_1(x)=f_2(x)=f_3(x)=0\}$ is a two-dimensional differentiable submanifold of $\Bbb{R}^4$.
It would suffice to show that the Jacobian of $F=(f_1,f_2,f_3)$ has rank $2$ for any point of $M$, right? I've been playing around with it for a while, but I'm not sure why $$DF(x)=\begin{bmatrix}x_3&-2x_2&x_1&0\\0&x_4&-2x_3&x_2\\x_4&-x_3&-x_2&x_1\end{bmatrix}$$ should always have rank $2$ for $x=(x_1,...x_4)\in M$. Is there a better way to do this?
Note that, when $f_1=0= f_2$ and $x_i\neq 0 \forall i$, the equation $f_3=0$ is automatically fulfilled (by calculation).
With this you may restrict your considerations to $f_1$ and $f_2$ and, to complete the reasoning, to points where $x_i=0$ for some $i$.
I'll leave the details to you, though.