Answer
The set $W$ is a subspace of $V$.
Work Step by Step
Let $u_{1}=a_{1}x+b_{1}y+c _{1}z\in W$ and $u_{2}=a_{2}x+b_{2}y+c_{2} z\in W$ and $\alpha , \beta \in W$ ,
1) $u_{1}+u_{2}=(a_{1}+a_{2})x+(b_{1}+b_{2})y+(c_{1}+c_{2})z $
since $a_{1}, a_{2} , b_{1} , b_{2} , c_{1} , c_{2}$ are scalars, then $u_{1}+u_{2}\in W$
2) $\alpha u_{1} =\alpha a_{1}x+\alpha b_{1}y+ \alpha c _{1}z\in W$
where $\alpha a_{1} , \alpha b_{1} , \alpha c_{1} $ are scalars
3) THERE exist $0 \in W$ is $0_{v}$ such that $ a_{1} . 0_{v}+b_{1} .0_{v} +c _{1}.0_{v}=0_{v} $
4) for all element $u_{1} \in W$ , there is its inverse in the form
$-u_{1}=-a_{1}x+(-b_{1})y+(-c _{1})z\in W $,
Because -a_{1} , (-b_{1}) , (-c _{1}) are scalars.
Thus the set $W$ is a subspace of $V$.
