I was told the following
$(1,-1,0)$ induces the specific embedding of $\text{SL}_2(\Bbb C)$ into $\text{SL}_3(\Bbb C)$: $$\begin{bmatrix}a&b&0\\c&d&0\\0&0&1\end{bmatrix}$$ and $(0,1,-1)$ induces: $$\begin{bmatrix}1&0&0\\0&a&b\\0&c&d\end{bmatrix}$$
How is this true? I don't even know how to start to figure out what this means.
Without more context it's difficult to be certain what meaning was intended, but I conjecture that $(1,-1,0)$ and $(0,1,-1)$ are to be interpreted as root vectors in the root system of $SL_3(\mathbb C)$. Let $\mathfrak h$ be the usual Cartan subalgebra of $\mathfrak{sl}_3(\mathbb C)$, namely containing trace zero diagonal matrices. Let $r=(1,-1,0)$. Then $e=E_{1,2}$ is in the $r$-root space, in the sense that $$ [h,e]=\mathrm{tr}(h\,\mathrm{diag}(r))e $$ for all $h\in\mathfrak h$. Similarly $f=E_{2,1}$ is in the $(-r)$-root space. The Lie subalgebra of $\mathfrak{sl}_3(\mathbb C)$ generated by $e$ and $f$ consists of trace zero matrices of the form $$ \begin{pmatrix}a&b&0\\ c&d&0\\ 0&0&0\end{pmatrix}. $$ Applying the matrix exponential, the corresponding Lie subgroup of $SL_3(\mathbb C)$ is the first one you describe. Similarly for the other root.