Let $ M=\{[z_0,z_1,z_2, z_3] \in \mathbb{RP}^3 | (z_0-z_3)^2+az_1^2=0\}$, where $a\in \mathbb{R}$. Show that $M$ is a smooth submanifold of $\mathbb{RP}^3$ of dimension $2$ when $a=0$, but not if $a\ne0.$
So far:
If $a=0 \implies z_0=z_3$, so we are looking at the subspace $N\subset M$, where $N=\{[z_0,z_1,z_2, z_0] \in \mathbb{RP}^3 \}\cong \{[z_0,z_1,z_2] \} \cong \mathbb{RP}^2$, and $\mathbb{RP}^2$ is a smooth manifold of dimension 2.
If $a\ne 0$, I'm not really sure how to proceed. I tried to do as I did above and got $N=\{[z_0, \pm (-a)^{-1/2}(z_0 -z_3),z_2, z_3] \in \mathbb{RP}^3 \}$.
Note: I initially tried using parameterizations and Regular Value Thm for the first part, but 0 is not a regular value.
If you take the case where $a$ is negative, just say $-1$ for ease, you get, $$z_0-z_3+z_1=0$$ or $$z_0-z_3-z_1=0$$ the union of two planes, so their points of intersection are not smooth.
If $a$ is positive, say $1$, then you have $$z_0=z_3$$and $$z_1=0$$ this seems to be the intersection and its a line, so it seems to me to be smooth ? But its dimension is $1$, so maybe this is what was meant.