How to calculate $a,b,c,d$ given the eigenvectors 
Given the matrix 
  $$A=\begin{bmatrix}
        a & b & 2 \\
        c & d & 6 \\
        3 & 4 & -3 
        \end{bmatrix}$$
  with  eigenvectors
  $$v_1=\begin{bmatrix}
        5 \\
        1 \\
        3 
        \end{bmatrix}
\quad\text{and}\quad v_2=\begin{bmatrix}
        7 \\
        4 \\
        3 
        \end{bmatrix}$$
  find $a,b,c,d$.

After this i know i should compute $Av_1$ and $Av_2$.
$$Av_1=\begin{bmatrix}
        5a+b+6 \\
        5c+d+18 \\
        10
        \end{bmatrix}$$
So what next in order to find value for $a,b,c,d$?
 A: $A*V_1 = \lambda*V_1$
$A*V_1$ is
\begin{bmatrix}
        5a+b+6 \\
        5c+d+18 \\
        10 \\
        \end{bmatrix}
$\lambda*V_1$ is
\begin{bmatrix}
        5*\lambda \\
        1*\lambda \\
        3*\lambda \\
        \end{bmatrix}
On equating, we see that $\lambda = \frac{10}{3}$
Similarly solve for other eigen value. Then with equations formed, get the desired values
A: Hint. Note that by the definitions of eigenvalue and eigenvector,
$$\begin{bmatrix}
        5a+b+6 \\
        5c+d+18 \\
        10 
        \end{bmatrix}=Av_1=\lambda_1 v_1=\begin{bmatrix}
        5\lambda_1 \\
        \lambda_1 \\
        3\lambda_1 
        \end{bmatrix}$$ 
    and looking at the third component wefind that $10=3\lambda_1$, that is $\lambda_1=10/3$. Hence
$$\begin{cases}
5a+b+6=5\lambda_1 \\
5c+d+18=\lambda_1  \\
\end{cases}\implies \begin{cases}
5a+b=32/3 &\text{(1)}\\
5c+d=-44/3 &\text{(2)}\\
\end{cases}.$$
Now do the same thing with the other eigenvector $v_2$ and you will obtain another couple of equations (3) and (4). Finally solve the linear system of (1) and (3) with respect to $a$, $b$ and the linear system of (2) and (4) with respect to $c$, $d$. 
