I'm trying to solve this question by using the dot product property for vectors that are perpendicular where if you multiply the vectors, they should equal $0$. I've tried the following:
- $ 20 + 3y + 4z = 0$
- $3y + 4z = -20$
I'm unsure how to find a $y$ or $z$ value that makes $(10, y, z)$ perpendicular to $(2, 3, 4)$ from here.
You're basically done. For any pair $(y, z)$ that satisfies the linear relation $3y + 4z = -20$, the vector $(10, y, z)$ is perpendicular to $(2, 3, 4)$. For example $(y, z) = (-4, -2)$ works and you can check that $(10, -4, -2)$ works. Here's a picture of the linear relation:
I've indicated some points with integer coordinates, but any points on the line work. And each one corresponds to a vector perpendicular to $(2, 3, 4)$.
There's a whole line's worth of solutions because in $3$-dimensional space, there's a whole $2$-d plane's worth of vectors perpendicular to a give one. Once we restrict our attention to only those vectors in the plane for which the first coordinate $x=10$, then we get the $1$-d line of solutions. In general$^\dagger$ each equation cuts the dimension of the space of solutions down by $1$. Here we have $2$ equations in a $3$-d space and $3-2=1$.
$^\dagger\!$ Lines, planes, and other subspaces can be parallel or otherwise not intersect in this generic fashion, but these are interesting edge cases. This discussion of dimension is only meant to inspire intuition about dimensions, and it is exactly what happens in all but those edge cases.