1. **State the problem:** We need to show that the vectors $\mathbf{v_1} = (3,0,-3)$, $\mathbf{v_2} = (-1,1,2)$, $\mathbf{v_3} = (4,2,-2)$, and $\mathbf{v_4} = (2,1,1)$ are linearly dependent over $\mathbb{R}$. This means there exist scalars $a,b,c,d$, not all zero, such that: $$a\mathbf{v_1} + b\mathbf{v_2} + c\mathbf{v_3} + d\mathbf{v_4} = \mathbf{0}$$ 2. **Set up the equation:** $$a(3,0,-3) + b(-1,1,2) + c(4,2,-2) + d(2,1,1) = (0,0,0)$$ This gives the system of equations: $$\begin{cases} 3a - b + 4c + 2d = 0 \\ 0a + b + 2c + d = 0 \\ -3a + 2b - 2c + d = 0 \end{cases}$$ 3. **Solve the system:** From the second equation: $$b = -2c - d$$ Substitute $b$ into the first and third equations: First: $$3a - (-2c - d) + 4c + 2d = 0 \implies 3a + 2c + d + 4c + 2d = 0$$ Simplify: $$3a + 6c + 3d = 0$$ Third: $$-3a + 2(-2c - d) - 2c + d = 0 \implies -3a - 4c - 2d - 2c + d = 0$$ Simplify: $$-3a - 6c - d = 0$$ 4. **Add the two simplified equations:** $$ (3a + 6c + 3d) + (-3a - 6c - d) = 0 + 0 \implies 2d = 0 \implies d = 0$$ 5. **Substitute $d=0$ back:** From first simplified: $$3a + 6c = 0 \implies 3a = -6c \implies a = -2c$$ From third simplified: $$-3a - 6c = 0$$ Substitute $a = -2c$: $$-3(-2c) - 6c = 6c - 6c = 0$$ True for all $c$. 6. **Recall $b = -2c - d = -2c$ since $d=0$.** 7. **Choose $c=1$ for a nontrivial solution:** $$a = -2, b = -2, c = 1, d = 0$$ 8. **Verify:** $$-2(3,0,-3) + (-2)(-1,1,2) + 1(4,2,-2) + 0(2,1,1) = (-6,0,6) + (2,-2,-4) + (4,2,-2) + (0,0,0) = (0,0,0)$$ 9. **Conclusion:** Since a nontrivial linear combination of the vectors equals zero, the vectors are linearly dependent over $\mathbb{R}$.