Help with finding the vectors $\vec{a_1}$ and $\vec{a_2}$ So I have the following problem:

So I'm thinking this:
$\vec{a}=2\vec{e_1}+\vec{e_2}+\vec{e_3}=\begin{bmatrix}2\\1\\1\end{bmatrix}$
$\vec{b}=3\vec{e_1}+2\vec{e_2}+5\vec{e_3}=\begin{bmatrix}3\\2\\5\end{bmatrix}$
If we have a vector $\vec{a_1}$ then we must have that $\vec{a_1}=d\vec{b}=d\begin{bmatrix}3\\2\\5\end{bmatrix}$
For the vector $\vec{a_2}=\begin{bmatrix}x\\y\\z\end{bmatrix}$ we must have that
$\vec{a_2}\bullet\vec{b}=\begin{bmatrix}x\\y\\z\end{bmatrix}\bullet\begin{bmatrix}3\\2\\5\end{bmatrix}=0$
$\vec{a_2}\bullet\vec{b}=3x+2y+5z=0$
From this I get that:
$\vec{a_1}+\vec{a_2}=d\begin{bmatrix}3\\2\\5\end{bmatrix}+\begin{bmatrix}\frac{-2y-5z}{3}\\\frac{-3x-5z}{2}\\\frac{-3x-2y}{5}\end{bmatrix} =\begin{bmatrix}2\\1\\1\end{bmatrix}$
So I get three equations with 4 unknowns, that I'm not able to solve. I'm really stuck. Is there another way to solve this problem, or I'm I missing something?

You have $3x+2y+5z = 0\Rightarrow y = -\frac{3x+5z}{2}$
Substitute only one variable (here $y$ in terms of $x$ and $z$), this will reduce the no. of variables by $1$.
The mistake you've done is you've substituted $x$ in terms of $y,z$, $y$ in terms of $z,x$ and $z$ in terms of $x,y$, so the no. of variables to be determined remain same.
Now,
$$\begin{align}&3d+x = 2 \cdots (\text{eq. }1)   
\\&2d- \frac{3x+5z}{2}=1\cdots (\text{eq. }2) 
\\&5d+z = 1 \cdots (\text{eq. }3)\end{align}$$
Perform $3\times(\text{eq. }1)+2\times(\text{eq. }2)+5\times(\text{eq. }3)$ to get $d$. Then you can find $x,z$ and finally $y$
