1. **Problem statement:** Determine whether the vector $w = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}$ is in the range of the linear operator $T : \mathbb{R}^3 \to \mathbb{R}^3$ defined by $$ T(v_1) = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}, \quad T(v_2) = \begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix}, \quad T(v_3) = \begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix}. $$ 2. **Recall:** The range $R(T)$ is the span of $\{T(v_1), T(v_2), T(v_3)\}$. To check if $w$ is in $R(T)$, we need to see if there exist scalars $a,b,c$ such that $$ a T(v_1) + b T(v_2) + c T(v_3) = w. $$ 3. **Set up the system:** $$ a \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + b \begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix} + c \begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}. $$ This gives the system of equations: $$ \begin{cases} - a + 0 b - c = -2 \\ 2 a + 5 b - c = 1 \\ a + 0 b + 2 c = 2 \end{cases} $$ 4. **Rewrite:** $$ \begin{cases} - a - c = -2 \\ 2 a + 5 b - c = 1 \\ a + 2 c = 2 \end{cases} $$ 5. **From the first equation:** $$ - a - c = -2 \implies a + c = 2. $$ 6. **From the third equation:** $$ a + 2 c = 2. $$ 7. **Subtract the first from the third:** $$ (a + 2 c) - (a + c) = 2 - 2 \implies c = 0. $$ 8. **Since $c=0$, from $a + c = 2$ we get:** $$ a = 2. $$ 9. **Substitute $a=2$, $c=0$ into the second equation:** $$ 2(2) + 5 b - 0 = 1 \implies 4 + 5 b = 1 \implies 5 b = -3 \implies b = -\frac{3}{5}. $$ 10. **Conclusion:** There exist scalars $a=2$, $b=-\frac{3}{5}$, $c=0$ such that $$ 2 T(v_1) - \frac{3}{5} T(v_2) + 0 \cdot T(v_3) = w. $$ Therefore, $w$ is in the range of $T$. --- 11. **Find a basis for $R(T)$:** The vectors $T(v_1), T(v_2), T(v_3)$ are $$ \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}, \quad \begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix}. $$ Check if they are linearly independent. 12. **Form matrix with these as columns:** $$ A = \begin{pmatrix} -1 & 0 & -1 \\ 2 & 5 & -1 \\ 1 & 0 & 2 \end{pmatrix}. $$ 13. **Check linear dependence:** Suppose $x,y,z$ satisfy $$ -1 x + 0 y -1 z = 0 \\ 2 x + 5 y -1 z = 0 \\ 1 x + 0 y + 2 z = 0. $$ 14. **From first:** $$ - x - z = 0 \implies z = -x. $$ 15. **From third:** $$ x + 2 z = 0 \implies x + 2(-x) = 0 \implies -x = 0 \implies x = 0. $$ 16. **Then $z = -x = 0$. Substitute into second:** $$ 2(0) + 5 y - 0 = 0 \implies 5 y = 0 \implies y = 0. $$ 17. **Only trivial solution, so vectors are linearly independent.** 18. **Therefore, a basis for $R(T)$ is** $$ \left\{ \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix} \right\}. $$ --- 19. **Find dimension of null space $N(T)$:** By rank-nullity theorem, $$ \dim(\mathbb{R}^3) = \dim(R(T)) + \dim(N(T)) \implies 3 = 3 + \dim(N(T)) \implies \dim(N(T)) = 0. $$ 20. **Hence, the null space contains only the zero vector.**