1. **State the problem:** Show that the set $W = \{(x,y,z) \in \mathbb{R}^3 : x + y + z = 0\}$ is a subspace of $\mathbb{R}^3$. 2. **Recall the subspace criteria:** A subset $W$ of a vector space $V$ is a subspace if: 1. The zero vector is in $W$. 2. $W$ is closed under vector addition. 3. $W$ is closed under scalar multiplication. 3. **Check zero vector:** The zero vector in $\mathbb{R}^3$ is $(0,0,0)$. Check if it satisfies the condition: $$0 + 0 + 0 = 0,$$ so $(0,0,0) \in W$. 4. **Check closure under addition:** Let $u = (x_1,y_1,z_1)$ and $v = (x_2,y_2,z_2)$ be in $W$. Then: $$x_1 + y_1 + z_1 = 0 \quad \text{and} \quad x_2 + y_2 + z_2 = 0.$$ Consider $u + v = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$. Sum of components: $$ (x_1 + x_2) + (y_1 + y_2) + (z_1 + z_2) = (x_1 + y_1 + z_1) + (x_2 + y_2 + z_2) = 0 + 0 = 0,$$ so $u + v \in W$. 5. **Check closure under scalar multiplication:** Let $u = (x,y,z) \in W$ and $c \in \mathbb{R}$. Then: $$x + y + z = 0.$$ Consider $cu = (cx, cy, cz)$. Sum of components: $$cx + cy + cz = c(x + y + z) = c \cdot 0 = 0,$$ so $cu \in W$. 6. **Conclusion:** Since $W$ contains the zero vector, and is closed under addition and scalar multiplication, $W$ is a subspace of $\mathbb{R}^3$. **Final answer:** $W$ is a subspace of $\mathbb{R}^3$.