1. **Problem:** Determine if the set of all vectors of the form $(a,0,0)$ in $\mathbb{R}^3$ is a subspace. 2. **Subspace Test:** A subset $W$ of $\mathbb{R}^3$ is a subspace if it satisfies three conditions: 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 $(0,0,0)$. Since $(0,0,0)$ can be written as $(a,0,0)$ with $a=0$, the zero vector is in the set. 4. **Check closure under addition:** Take two vectors in the set: $(a,0,0)$ and $(b,0,0)$. Their sum is $(a+b,0+0,0+0) = (a+b,0,0)$, which is still in the set. 5. **Check closure under scalar multiplication:** Take a vector $(a,0,0)$ and scalar $k$. The product is $(ka,k\cdot0,k\cdot0) = (ka,0,0)$, which is in the set. 6. **Conclusion:** All three conditions are met, so the set of all vectors of the form $(a,0,0)$ is a subspace of $\mathbb{R}^3$. **Final answer:** Yes, the set is a subspace.