Find the condition on $a,b,c$ such that $(a,b,c) \in \operatorname{span}((2,1,0) , (1,-1,2), (0,3,-4))$ Find the condition on $a$, $b$, $c$  so that $(a,b,c) \in \mathbb{R^3}$ is in the space generated by $(2,1,0)$, $(1,-1,2)$ and $(0,3,-4)$
If $(a,b,c)$ lies in the space generated by given vectors then it must be equal to the linear span.
$(a,b,c) = \alpha(2,1,0) +\beta(1,-1,2) + \gamma(0,3,-4)$ 
Upon solving this system for $\alpha$, $\beta$ and $\gamma$  I ended up with following equations
$\alpha - \beta + 2\gamma = b$
$0\alpha + 3\beta - 4\gamma = a-2b$
$0\alpha + 0\beta + 0\gamma =  c-a+2b$
Since the given system needs to be consistent we must have $c-a+2b = 0$
However, my book says the correct answer is $2a - 4b - 3c = 0$
Please tell me what is wrong with my solution and what should be  the correct way to solve it.
Thank you.
 A: 
$(a,\color{blue}{b},c)$ = $\color{blue}{\alpha}(2,\color{blue}{1},0)$ +$\color{blue}{\beta}(1,\color{blue}{-1},2)$ + $\color{blue}{\gamma}(0,\color{blue}{3},-4)$
Upon solving this system for $\alpha$ , $\beta$ and $\gamma$  I ended up with following equations
$\alpha$ - $\beta$ + $\color{red}{2}\gamma$ = $b$

How did you arrive at the equation above? Looking at the second coordinate (in blue); I'd expect:
$$b=\alpha-\beta+\color{purple}{3}\gamma$$
A: Your book is right. The system is$$\left\{\begin{array}{l}2\alpha+\beta=a\\ \alpha-\beta+3\gamma=b\\ 2\beta-4\gamma=c.\end{array}\right.$$Multiplying the second equation by $-2$ and adding this to the first one, you get$$\left\{\begin{array}{l}3\beta-6\gamma=a-2b\\ \alpha-\beta+3\gamma=b\\ 2\beta-4\gamma=c.\end{array}\right.$$This is equivalent to$$\left\{\begin{array}{l}\beta-2\gamma=\frac13(a-2b)\\ \alpha-\beta+3\gamma=b\\ \beta-2\gamma=\frac12c\end{array}\right.$$So, the system is consistent if and only if $\frac13(a-2b)=\frac12c$, which is equivalent to the answer provided by your book.
A: Let us try to find $c_1$, $c_2$, $c_3$ such that the given vectors fulfill the condition $c_1x_1+c_2x_2+c_3x_3=0$. (If this is true, any vector $(a,b,c)$ belonging to their span fulfills the same condition.)
From this we get a system of three linear equations
\begin{align*}
2c_1+c_2&=0\\
c_1-c_2+2c_3&=0\\
c_1+2c_2-2c_3&=0
\end{align*}
Solving this system of equations we get that $(c_1,c_2,c_3)$ has to be a multiple of the vector $(2,4,3)$. 
This can be said equivalently like this: The span of the three given vectors is exactly the set of vectors $\{(x_1,x_2,x_3)\in\mathbb R^3; 2x_1+4x_2+3x_3=0\}$.

This is related to the fact that every subspace of $\mathbb R^n$ is equal to solution set of some homogeneous system of linear equations.
By solving the system with unknowns $c_{1,2,3}$ we found such homogeneous system. (In this case it consisted of only one equation.) 
