Determining Basis of any vector equation with only one formula

Question

I'm really struggling with finding the basis when it's just like $ax + by + cz = d$ :

Answer 1

Note that, since you have an over-determined system with two free variables, you can take $y$ and $z$ as any constants. Say $y = c_{1}$ and $z = c_{2}$

In that case, all elements of the subspace $V$, can be written as $(2c_{1} - 3c_{2}, c_{1}, c_{2})$.

This can be re-written as $(2c_{1}, c_{1}, 0) + (-3c_{2}, 0, c_{2})$

Taking $c_1$ and $c_2$, common, we have $c_{1}(2,1,0) + c_{2}(-3,0,1)$

So, the vectors $\lbrace (2,1,0),(-3,0,1) \rbrace$ are Linearly Independent and span $V$. Hence, this forms a basis for $V$.

If $u \in$ span$(\lbrace (2,1,0),(-3,0,1) \rbrace)$ $\implies$ $u = c_{1}(2,1,0) + c_{2}(-3,0,1) = (2c_{1} - 3c_{2}, c_{1}, c_{2})$

$x - 2y + 3z = 0 \implies u \in V$

span$(\lbrace (2,1,0),(-3,0,1) \rbrace) \subset V$

Conversely, if $v = [x$ $y$ $z]^T$ $\in V$, then $x - 2y + 3z = 0$, which is how we came here.