Let us assume that we have a plane H given by:
$$ H=\lbrace\lambda\left\lbrack1,-1,0\right\rbrack^{^{^T}}+\alpha\left\lbrack0,-1,1\right\rbrack^{^{^T}}\mid\alpha,\lambda\in\mathbb{R}\rbrace $$
Is there a way to describe this plane as:
$$ Ax+By+Cy+D=0 $$
If so, what would be the most effective way of finding this plane? And if not, why is this not possible?
Porting my comment to an answer per request: since the vectors $[1,-1,0]^T$ and $[0,-1,1]^T$ are linearly independent in $\mathbb{R}^3$, you can compute their cross product to obtain the vector $[-1,-1,-1]^T$ which is normal to the plane spanned by the two vectors. A vector $[x,y,z]^T$ is in this plane if and only if $$[-1,-1,-1]\cdot[x,y,z]^T=-x-y-z=0$$ So you can take $A=B=C=-1$ and $D=0$ in the equation (where $Cy$ should be $Cz$).