We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$. 
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .

What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?

If you happen to know eigenvalues/eigenvectors, your issue can be written :
$$\underbrace{\begin{pmatrix}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{pmatrix}}_A\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}=r\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}$$
meaning that $r$ must be an eigenvalue of matrix $A$.
The eigenvalues of $A$ are $3$ and $-1$ ($-1$ being a
triple eigenvalue, but the text allows to take it only once) associated to resp. eigenvectors:
$$\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}=\begin{pmatrix}1\\1\\1\\1\\\end{pmatrix},\begin{pmatrix}1\\1\\1\\-3\\\end{pmatrix}$$
fulfilling the condition that neither $a$, nor $b,c,d$ are zero (being in the denominator of the initial equation)
Therefore the answer is $3+(-1)=2$.
