Can you find the result of a linear transformation by projecting a given vector onto known eigenvectors? I have a practice exercise I'm struggling with. Essentially, it gives you the eigenvalues and eigenvector for a given linear transformation and then asks you to find the result of the transformation on some other vector.
$$
A\begin{bmatrix}-1\\-1\\-2\end{bmatrix} = 1\begin{bmatrix}-1\\-1\\-2\end{bmatrix}
$$
$$
A\begin{bmatrix}1\\1\\1\end{bmatrix} = 0\begin{bmatrix}1\\1\\1\end{bmatrix}
$$
$$
A\begin{bmatrix}-1\\-4\\-3\end{bmatrix} = 2\begin{bmatrix}-1\\-4\\-3\end{bmatrix}
$$
Find $A\begin{bmatrix}3\\-4\\3\end{bmatrix}$.
My first thought is to split this vector into components along the known eigenvector, calculate the result, and then add the vectors back together, but I don't know if this is valid. If it isn't, how would one go about this problem? 
Thanks for any and all help.

Yes, this is the right approach; the idea is to appeal to linearity. First decompose the given vector into a linear combination of the eigenvectors, then:
$$\begin{align}
A\left(c_1\begin{bmatrix}-1\\-1\\-2\end{bmatrix} + c_2\begin{bmatrix}1\\1\\1\end{bmatrix} + c_3\begin{bmatrix}-1\\-4\\-3\end{bmatrix}\right) &= c_1\left(A\begin{bmatrix}-1\\-1\\-2\end{bmatrix}\right) + c_2\left(A\begin{bmatrix}1\\1\\1\end{bmatrix}\right) + c_3\left(A\begin{bmatrix}-1\\-4\\-3\end{bmatrix}\right)\\
&=c_1\left(1\begin{bmatrix}-1\\-1\\-2\end{bmatrix}\right) + c_2\left(0\begin{bmatrix}1\\1\\1\end{bmatrix}\right) + c_3\left(2\begin{bmatrix}-1\\-4\\-3\end{bmatrix}\right)\\
&=c_1\begin{bmatrix}-1\\-1\\-2\end{bmatrix} + 2c_3\begin{bmatrix}-1\\-4\\-3\end{bmatrix}
\end{align}
$$
