A problem from Hardy's_Pure Mathematics on Real variables about Linear equations in 3 variables It may be my stupidity, yet I am unable to understand what the following problem asks
(Q.1 Miscellaneous Examples, Chapter-1: Real Variables).
It says: 

What are the conditions that $ax+by+cz=0$, 
  
  
*
  
*(1) for all values of $x,y,z$  ; 
  
*(2) for all values of $x,y,z$ subject to $\alpha x+\beta y+\gamma z=0$; 
  
*(3) for all values of $x,y,z$ subject to both $\alpha x+\beta y+\gamma z=0$ and $Ax+By+Cz=0$?" 
  

Initially I thought it was about linear diophantine equations in three variables, but then it is already given that the constant term is $0$ which means that it is a solvable diophantine equation (for all the coefficients divide $0$). So the problem is surely not about the solvability of the equation. Then, what it is about? What does the author mean by 'conditions'? Any suggestions are welcome.
 A: I think Hardy is asking for conditions on $(a,b,c)$, to be given in terms of $(A,B,C)$ and $(\alpha,\beta,\gamma)$. The $(x,y,z)$ are dummy variables, note the "for all".
For (1), which $(a,b,c)$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $(a,b,c)=(0,0,0)$. 
For (2), which $(a,b,c)$ are orthogonal to all vectors which are orthogonal to $(\alpha,\beta,\gamma)$? Answer, clearly those which are parallel to $(\alpha,\beta,\gamma)$. So the condition is that for some scalar $\lambda$, we must have $(a,b,c)=\lambda (\alpha, \beta, \gamma)$.
Case (3) is left to the reader; the condition is clearly that for some scalars $\lambda, \mu$ we must have that $(a,b,c)=\lambda (\alpha, \beta, \gamma)+\mu (A,B,C)$.  
A: 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.
