Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $1$. It should likely help give you an idea.
However, here is what it seems to me. As it's a book about real variables, I doubt the equations are supposed to be diophantine ones, i.e., where the variables can only be integers. Instead, it's to do with the degrees of freedom vs. how many linearly independent constraints you have. With just $1$ equation, you can generally pick any $2$ values and then determine the third, e.g., choose $y$ and $z$ to then get $x$ from
$$x = \frac{-by-cz}{a} \tag{1}\label{eq1}$$
This is assuming that $a \neq 0$.
With $2$ constraint equations, assuming they're linearly independent, you can solve each for $1$ variable to eliminate it, resulting in an equation with $2$ variables. You then have $1$ degree of freedom to choose $1$ of those remaining variables, with the other $2$ then being determined. For example, using your provided $2$ equations:
$$ax + by + cz = 0 \tag{2}\label{eq2}$$
$$\alpha x + \beta y + \gamma z = 0 \tag{3}\label{eq3}$$
Multiply \eqref{eq2} by $\alpha$, multiply \eqref{eq3} by $a$, and subtract the $2$ resulting equations to get
$$\left(\alpha b - a\beta\right)y + \left(\alpha c - a\gamma\right)z = 0 \; \Rightarrow \; y = \left(\frac{a\gamma - \alpha c}{\alpha b - a\beta}\right)z \tag{4}\label{eq4}$$
This is assuming that $\alpha b - a\beta \neq 0$. Thus, for each $z$, you get a $y$ from \eqref{eq4} and then an $x$ from \eqref{eq1}.
For $3$ linear constraint equations, assuming they're all linearly independent, you get just one set of values.
If any of these equations are linearly dependent, this results in an extra degree of freedom for each equation that is dependent.
Based on your understanding of chapter 1, please indicate if this doesn't match what you might expect.
Update: Based on the comment by ancientmathematician about a geometric argument, note that one equation gives a plane, having $2$ independent equations gives the intersection of $2$ planes, i.e., a line, and $3$ independent equations gives the intersection of $3$ planes, i.e., a point. Perhaps these are the "conditions" being referred to. However, based on how the question is specifically worded, I think a better fit would be what is described in ancientmathematician's answer.