Let $X\subseteq R^3$ That defined by : $X=\{(x_1,x_2,x_3)|x_1^2+x_2^2-x_3^2=1\}$ Find $Sp(X)$, dim(X), and find basis for $X$

I have this problem.

Let $X\subseteq R^3$ That defined by : $$X=\{(x_1,x_2,x_3)|x_1^2+x_2^2-x_3^2=1\}$$

Find $Sp(X)$, dim(X), and find basis for $X$

My solution :

Let $x_1,x_2 \in R$

$$x_1^2+x_2^2-x_3^2=1 \implies x_1^2+x_2^2-1=x_3^2$$

$$[B]_u= (x_1^2,x_2^2,x_1^2+x_2^2-1)$$ $$dim(X)=2$$ $$X = Sp\{(x_1^2,x_2^2,x_1^2+x_2^2-1)\} = Sp\{x_1^2(1,0,1-1)+x_2^2(0,1,1-1)\}=Sp(\{(1,0,0),(0,1,0)\}$$

But $Sp(\{(1,0,0),(0,1,0)\}$ is incorrect since (for exmaple) $(2,0,0) \notin X$

I pretty sure the rest is correct, Any idea how to get the span of X?

Thank you!

The set $X$ that you've described is not a subspace of $\mathbb R^3$, so it doesn't have a basis. For instance, $(1, 0, 0) \in X$, but $(2, 0, 0) \notin X$, showing that it's not closed under scalar multiplication.
On the other hand, we have $(1,0,0) \in X$, $(0, 1, 0) \in X$ and $(1, 1, 1) \in X$, from which we can conclude that $(0,0,1)$ is in the span of $X$, hence all of $\mathbb R^3$ is in $span(X)$, so $span(X) = \mathbb R^3$.
Because $X$ is not a subspace, it does not have a dimension (as a subspace). As a manifold, it has dimension 2. But I doubt that this is relevant to your question.