Value of an element of a symmetric matrix with given eigenvalue. Let \begin{bmatrix}
    a & b &  c \\
    b & d & e \\
    c & e & f
\end{bmatrix} be a real matrix with eigenvalues $1$,$0$ and $3$. If the eigenvectors corresponding to $1$ and $0$ are $(1,1,1)^{T}$ and$(1,-1,0)^{T}$ respectively, then what is the value of $3f?$  How to find value of $3f?$
From the given conditions on eigenvalues we have $$a+b+c=1\\b+d+e=1\\c+e+f=1\\a-b=0\\b-d=0\\c-e=0.$$ Is there any short way to find value of $f?$ Please suggest me. Thanks a lot.

We know trace $a+d+f=4$
Take help from equations you have got from $(A-\lambda I)X=0$
$a=b=d$ and $c=e$
Now $a+d+f=2a+f=4-----(1)$
Find $a$ which is not very hard 
$a+b+c=2a+c=1------(2)$
$c+e+f=2c+f=1------(3)$
Solving $(2),(3)$ we have $4a-f=1----(4)$
Solving $1$ and $4$ we have $f=\dfrac{7}{3}$
$3f=7$
