how do you find the intersection of planes.
For example,
-2x+3y-4z=3
4x +2y+5z=5
x+6y+3z=5
am having trouble getting my head around it?
Wouldn't you just find the $\displaystyle (x, y, z)$ that satisfies all 3 equations?
For convenience in calculations, I'll switch the first and third equations.
$\displaystyle x + 6y + 3z = 5$
$\displaystyle 4x + 2y + 5z = 5$
$\displaystyle -2x + 3y - 4z = 3$
$\displaystyle R_2 - 4R_1 \to R_2$ and $\displaystyle R_3 + 2R_1 \to R_3$
$\displaystyle x + 6y + 3z = 5$
$\displaystyle -22y - 7z = -15$
$\displaystyle 15y + 2z = 13$
Can you go from here?