What happens to the $d$ part of the plane equation $x+y+z=d$ when preforming a vector projection onto the plane? So i am working with vector projection and given a vector $v=(3,4,0)$ and a plane $x+y+z=1$ i am supposed to find the projection of $v$ onto the plane. The logical step is to just use the normal vector of the plane $n$ and do a projection $proj_n v=(v*n)/(n*n)*n$. But when i draw up the vectors and the plane, the projection is not onto the plane because of the $D$ part of the plane equation where $D=1$. 

Is this the right way to solve these kinds of problems or if not what do i need to do solve it?
The easiest way to find the projection of a vector $v$ to a plane with normal $n$ is to think of  the line $g(t)=v+t\cdot n$ and compute the intersection with your given plane.

As you are given the plane in coordinate form, we can directly see the normal vector to be $n=(1,1,1)$ and we know $v=(3,4,0)$.

Putting it all in the coordinate form equation, we get
$$ 
3+t+4+t+0+t=1\\
\Leftrightarrow 7+3t =1\\
\Leftrightarrow t = -2
$$

Thus, the projection onto the plane is
$$
g(-2)=(3,4,0)-2\cdot(1,1,1)=(1,2,-1)
$$

A short check shows that the coordinate form fulfills 
$$ 1+2-1=1,$$ so this is really inside the plane :)
$\textrm{proj}_n v = \frac{v\cdot n}{n \cdot n} n$ is the component of the vector normal to the plane. So the component of the vector parallel to the plane is $v - \textrm{proj}_n v$, or, in this case, $w=\left( \frac{2}{3},\frac{5}{3},-\frac{7}{3}\right)$.

The value of $d$ doesn't come into it regardless, because we only care about the direction of the vector and the plane. But you can check your work by taking a random point $P$ in the plane, e.g., $(1,1,-1)$, and verifying that $P + w$ is also in the plane.

You can also verify that $w \cdot \textrm{proj}_n v = 0$ and that $w + \textrm{proj}_n v = v$.