Is $U=\{(r,s,t)|r,s,t \in \mathbb{R}, -r+3s+2t=0\}$ a subspace of $\mathbb{R}^3$? 
Is $U=\{(r,s,t)|r,s,t \in \mathbb{R}, -r+3s+2t=0\}$ a subspace of $\mathbb{R}^3$?

So far all I know is that the zero vector is in the subspace. How would I go about checking if it is closed under scalar multiplication and addition?
I usually try a few examples to make sure that the three conditions for a subspace hold, but the extra condition at the end is throwing me off.

Here's a solution: suppose we have $\vec x = (r,s,t)\in U, \vec y = (r',s',t')\in U$, and $k \in \mathbb R$.  Now, consider the vector 
$$
\vec v = \vec x + k \vec y = (r+kr',s+ks',t+kt')
$$
How can we show that $v$ is a member of $U$?

 Hint: note that $$-(r+kr')+3(s+ks')+2(t+kt') = (-r+3s+2t)+k(-r'+3s'+2t')$$

Now, plug in $k=1$ to get closure under addition and $\vec x = \vec 0$ to get closure under scalar multiplication.
