The system of three DE I want to solve the following system of DE:
$$
\begin{cases} \dot{x} = 2x+6y -15z,  \\ \dot{y} =x+y-5z,\\ \dot{z} = x+2y-6z, \end{cases}
$$
First, I rewtite the coefficents in the matrix form:
$$A = \begin{bmatrix}
2 & 6 & -15\\
1 & 1 & -5\\
1 & 2 & -6
\end{bmatrix}$$Then, I find $$det(A-\lambda I) = \begin{vmatrix}
2 -\lambda& 6 & -15\\
1 & 1 -\lambda& -5\\
1 & 2 & -6-\lambda
\end{vmatrix}
=-(\lambda+1)^3
$$
$\lambda=-1$ is of multiplicity $3$ and I don't know how to continue.

When the coefficient matrix $A$ has only one (repeated) eigenvalue $\lambda$, you’re in luck: the exponential $e^{tA}$ is easily computed without having to find any eigenvectors, generalized or otherwise. If the eigenvalue’s algebraic and geometric multiplicities are equal, then it must be a multiple of the identity matrix, and the exponential is trivially $e^{\lambda t}I$. Otherwise, for a $3\times3$ matrix, $A-\lambda I$ is nilpotent of index at most $3$. Moreover, $\lambda I$ and $A-\lambda I$ commute, therefore $$e^{tA} = e^{\lambda t}e^{t(A-\lambda I)} = e^{\lambda t}\left(I+t(A-\lambda I)+\frac{t^2}2(A-\lambda I)^2\right).$$ You can save yourself a bit of work by examining $A-\lambda I$: it will be obvious if this is a rank-1 matrix, in which case $(A-\lambda I)^2=0$.  
In this case, you’ve found that $\lambda = -1$. We then have $$A-\lambda I = \begin{bmatrix}3&6&-15\\1&2&-5\\1&2&-5\end{bmatrix},$$ which is clearly a rank-one matrix. Therefore, $$e^{tA} = e^{-t}\begin{bmatrix} 1+3t & 6t &-15t \\ t & 1+2t & -5t \\ t & 2t & 1-5t \end{bmatrix}.$$ The general solution to the system of differential equations is then obtained by multiplying this matrix by a vector of arbitrary constants.  
It’s likely, though, that you’re meant to compute the Jordan decomposition of $A$ and use that to produce the solution to the system. This is a tedious and unnecessary process for this particular matrix.
