1. **State the problem:** We need to find the dimension of the subspace $S = \{(2a + 4b - c,\ 3a - 6b,\ -4a + b + 3c,\ -a + 3c) : a,b,c \in \mathbb{R}\}$, which is a subset of $\mathbb{R}^4$. 2. **Express $S$ as a span:** We can write any vector in $S$ as a linear combination of vectors with coefficients $a, b$, and $c$: $$ \begin{aligned} (2a + 4b - c,\ 3a - 6b,\ -4a + b + 3c,\ -a + 3c) &= a(2, 3, -4, -1) + b(4, -6, 1, 0) + c(-1, 0, 3, 3). \end{aligned} $$ 3. **Identify spanning vectors:** The vectors are $$ v_1 = (2, 3, -4, -1),\quad v_2 = (4, -6, 1, 0),\quad v_3 = (-1, 0, 3, 3). $$ 4. **Check linear independence:** Determine if these vectors are linearly independent. Consider equation $$ \alpha v_1 + \beta v_2 + \gamma v_3 = 0. $$ This gives system $$ \begin{cases} 2\alpha + 4\beta - \gamma = 0 \\ 3\alpha - 6\beta + 0\gamma = 0 \\ -4\alpha + \beta + 3\gamma = 0 \\ -\alpha + 0\beta + 3\gamma = 0 \end{cases} $$ 5. **Solve system:** From last equation, $-\alpha + 3\gamma = 0 \implies \alpha = 3\gamma$. Substitute into third equation: $$ -4(3\gamma) + \beta + 3\gamma = -12\gamma + \beta + 3\gamma = \beta - 9\gamma = 0 \implies \beta = 9\gamma. $$ Substitute into second equation: $$ 3(3\gamma) - 6(9\gamma) = 9\gamma - 54\gamma = -45\gamma = 0 \implies \gamma = 0. $$ Then $\gamma =0$, so $\alpha=0$ and $\beta=0$. 6. **Conclusion:** Only trivial solution exists $\Rightarrow v_1, v_2, v_3$ are linearly independent. 7. **Dimension:** Therefore, the dimension of $S$ is the number of linearly independent vectors, which is $3$. **Final answer:** $$\dim(S) = 3.$$