He found the cross product of the first two equations.
You can google cross product and see how it is worked out. Or just ask it here
Of course, a less "sophisticated" way to solve the equations would be:
Add the first and third equations to get .
That last equation is obviously equivalent to . Replacing with in the second equation gives for all values of . Putting for in the first equation gives so that .
That is , for any value of .
Since and depend linearly on the single parameter , that is a line in three space. Of course, if we multiply all of , , and by 4 (chosen for the "4" under the in the original given solution) we have is also on that line. And, taking as parameter, so that , we have , .
Then