Find all solutions to $2 x + 3 y + 4 z = 10$ I do not have a background in math, and am wondering what type of question this is.  I looked combinatorics optimization, and the knapsack problem, but found the vocabulary too dense.  
The problem:
Given the following array A = [2,3,4], what is the best way to find all arrays of coefficients of A that give the product equal to 10.
Ex: 
A = [2,3,4]
B = [5,0,0]
C = A * B = [10,0,0]
Sum of C = 10

All Answers: [5,0,0], [3,0,1], [2,2,0], [0,2,1]
Just to restate, what is the best way to find the above answers (keeping in mind the target product could be large, like 100,000)? 
 A: Any vector on the following plane in $\mathbb{R}^3$ can be the answer
$$2x+3y+4z=10$$

A: If you want to find all nonnegative integer solutions, i.e.,
$$\{ (x,y,z) \in \mathbb N^3 \mid 2 x + 3 y + 4 z = 10 \}$$
then you can simply brute-force it. In Haskell,
λ filter (\(x,y,z)->2*x+3*y+4*z==10) [ (x,y,z) | x <- [0,1..5], y <- [0,1..4], z <- [0,1..3]] 
[(0,2,1),(1,0,2),(2,2,0),(3,0,1),(5,0,0)]

Hence, there are $5$ nonnegative integer solutions
$$\{ (0,2,1), (1,0,2), (2,2,0), (3,0,1), (5,0,0) \}$$
A much smarter way of finding all nonnegative integer solutions is to write $x = 5 - \frac{3}{2} y - 2 z$. Let
$$y = 2 m_1 \qquad \qquad z = m_2$$
where $m_1 \in \{0,1,2\}$ and $m_2 \in \{0,1,2,3\}$. Hence, the solution set can be parametrized as follows
$$\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 5\\ 0\\ 0\end{bmatrix} +  m_1 \begin{bmatrix} -3\\ 2\\ 0\end{bmatrix} + m_2 \begin{bmatrix} -2\\ 0\\ 1\end{bmatrix}$$
In Haskell,
λ filter (\(x,y,z)->x>=0 && y>=0 && z>=0) [ (5-3*m1-2*m2,2*m1,m2) | m1 <- [0,1..2], m2 <- [0,1..3] ]
[(5,0,0),(3,0,1),(1,0,2),(2,2,0),(0,2,1)]

We obtain the same $5$ solutions, though in a different order. However, we searched over a set of cardinality $12$, rather than one of cardinality $120$, as previously.
