Answer
See below
Work Step by Step
Assume that $u$ be the vector in $S$
then $u=(x,y,z)=(x,2x,3x)$
We can write set $S$ as $S=\{u\in R^3:u=(x,2x,3x),x\in R\}$
A1. Let $u=(t,2t,3t)\\
v=(s,2s,3s) \in S$
Obtain $u+v=(t,2t,3t)+(s,2s,3s)\\=(t+s,2t+2s,3t+3s)\\=(t+s,2(t+s),3(t+s))\\=(m,2m,3m)$
where $m=t+s$
Coordinates of sum $u+v$ are of the form of vectors from $S$, hence set $S$ is closed under addition.
A2. Let $u=(t,2t,3t) \in S$
and $k \in R$
Obtain: $ku=k(t,2t,3t)=(kt,2kt,3kt)=(n,2n,3n)$
where $n=kt$
Coordinates of product $ku$ are of the form of vectors from $S$, hence set $S$ is closed under scalar multiplication.
Since $S$ is nonempty subset of $R^3$ and $S$ is closed under addition and scalar multiplication, $S$ is subspace of $R^3$
