Gram schmidt, incorrect answer 
I projected y onto x giving me a vector y1. I then subtracted y1 from y which gave me a vector orthogonal to x, I called this vector y2. This vector was $y_2 = (-1, -7, -1)$. I then normalized x and y2 and entered them as my answers. But it says my answer for y2 is incorrect. 

You have the right procedure. But
$$\eqalign{
y_1 ={\rm proj}_x(y)&={x\cdot y\over x\cdot x} x \cr
&={ 4\cdot 1+3(-11/2)+1(-1/2)\over 4\cdot4+3\cdot 3+1\cdot1 }x\cr
&={-13\over26 }x\cr &={-1\over2}\Biggl(\matrix{4\cr3\cr1}\Biggr)\cr
&= \Biggl(\matrix{-2\cr-3/2\cr-1/2}\Biggr).
}
$$
So,
$$
y_2=  \Biggl(\matrix{1\cr-11/2\cr-1/2}\Biggr)  - \Biggl(\matrix{-2\cr-3/2\cr-1/2}\Biggr)=
\Biggl(\matrix{3\cr-4\cr0}\Biggr).$$
Here's the salient point I want to make: It's a good idea to check your work when you can. As a quick spot check, we have $$y_2\cdot x=3\cdot4+(-4)(3)+0\cdot1=0,$$ as it should (the $y_2$ of your calculation isn't orthogonal to $x$). 
