How do I determine whether or not this is a subspace of $\Bbb{F}^3$?
$\{ (x_1, x_2, x_3)\in\Bbb{F}^3: x_1 + 2x_2 + 3x_3 = 0 \}$
If I wanted to disprove this being a subspace, I would just need to find a case where some vector violates closure or that it doesn't have a 0 vector. But how would I prove that it doesn't violate closure for multiplication or addition for all values within its space? What approach should I take for this and other polynomials that have some kind of condition?
To disprove that it is a subspace it suffices to find an element in the set which violates at least one of the required properties that is
$1) \ \vec{0} \in W$
$2) \ \vec{v}+\vec{w} \in W$
$3) \ \vec{cv}\to c \cdot \vec{v} \ ,c \in \mathbb{F}$
Otherwise to check whether the set is a subspace we need to check that all the properties hold for any vector in the set.
For example, clearly property "$1)$" is satisfied indeed
$$(x_1,x_2,x_3)=(0,0,0)\implies 0+2\cdot 0+3\cdot 0=0$$
For property "$2)$" let consider $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$ in the set then note that
$$x_1+2x_2+3x_3+y_1+2y_2+3y_3=(x_1+y_1)+2(x_2+y_2)+3(x_3+y_3)=0$$
then also the sum is in the set.
For property "$3)$" let consider $(x_1,x_2,x_3)$ in the set then $\forall c$
$$x_1+2x_2+3x_3=0 \implies cx_1+2cx_2+3cx_3=0$$
then also any multiple is in the set.
We can't find any cunterexample in that case.