Determine whether $U$ is a subspace of $\mathbb{R}^3$ Determine whether $U$ is a subspace of $\mathbb{R}^3$
$$U = \{(r, s, t) \mid r, s, \text{ and }t \in \mathbb{R}, −r + 3s + 2t = 0\}.$$
I have never seen such a question how would i prove this?
 A: HINT
$V \subseteq \mathbb{R}^n$ is a subspace if


*

*$\vec{0} \in V$

*$V$ is closed under addition and constant multiplication; i.e. if $\vec{v},\vec{w} \in V$ and $a,b \in \mathbb{R}$ then $a\vec{v} + b \vec{w} \in V$.


Can you prove both?
UPDATE
Let $a,b \in \mathbb{R}$ and $\vec{v}, \vec{w} \in U$. Then there exist $s,t,S,T \in \mathbb{R}$ such that
$$
\vec{v} = \begin{pmatrix} 2s+2t \\ s \\ t \end{pmatrix}\\
\vec{w} = \begin{pmatrix} 2S+2T \\ S \\ T \end{pmatrix}
$$
So let $\vec{u} = a\vec{v} + b\vec{w}$. Can you find the correct values for $s,t$ to show $\vec{u} \in U$ as well?
A: Hint:
$$
-r+3s+2t=0
$$
is a plane, so it's a subspace of $\mathbb{R}^3$.
A: First, prove that (0,0,0) is an solution to the equation:
−r+3s+2t=0
-0+3.0+2.0=0
0=0 (true)
And than, prove that if (a,b,c) and (d,e,f) are solutions to the equation, m.(a,b,c) + n.(d,e,f) must to be a solution too (m and n real numbers):
(a,b,c) is a solution, so: 
−r+3s+2t=0
-a+3b+2c=0 (i)
(d,e,f) is a solution, so: 
−r+3s+2t=0
-d+3e+2f=0 (ii)
Well, note that if (i) and (ii) are true (and they are), (iii) and (iv) will be too:
-a+3b+2c=0 (i)
m.(-a+3b+2c)=m.0
m.(-a+3b+2c)=0 (iii)
and:
-d+3e+2f=0 (ii)
n.(-d+3e+2f)=n.0 (ii)
n.(-d+3e+2f)=0(iv)
Finally, (iii)+(iv) is true too:
m.(-a+3b+2c)=0 (iii)
+
n.(-d+3e+2f)=0(iv)
=
m.(-a+3b+2c) + n.(-d+3e+2f)=0+0 (iii)+(iv)
m.(-a+3b+2c) + n.(-d+3e+2f)=0 (iii)+(iv)
So, m.(a,b,c) + n.(d,e,f) is a solution too!  
