Finding Eigenvectors using Eigenvalues and Matrix $ (Alternative Method ?)$ 
Find the eigenvectors given the eigenvalues 1, 4 and 9:
\begin{pmatrix}1&-3&-3\\ -8&6&-3\\ 8&-2&7\end{pmatrix}

Normally I would use the row reduction method but I came across this method in the solutions. Is it a different way to solve or the same method in a different form?
Given Solutions:

Any explanation would be helpful. Thanks!
 A: You want to solve
$$(A-\lambda_iI)v_i=0$$
Or in other words you want to find the kernel of the matrix $M_i=A-\lambda_iI$. 
For the case that your $\lambda_i$ are distinct as in your example, the dimension of the Eigenspace of $v_i$ is 1 and thus also $\dim\ker M_i=1$. The matrix $M_i$ thus has two linear independent row vectors $m_i^1,m_i^2$. Any vector not in $\operatorname{span}(m_i^1,m_i^2)$ will be in $\ker M_i$. 
To construct such a vector, which will be the basis vector of $\ker M_i$, i.e. the Eigenvector $v_i$, you can find a vector orthogonal to $m_i^1$ and $m_i^2$. In your $\mathbb R^3$ example this can be accomplished through $v_i=m_i^1\times m_i^2$. 
This is what you calculate in your determinants. The cross product of two rows of $A-\lambda_i I$.
A: For $3\times 3$ matrix $A$ having $\lambda$ as an eigenvalue, you write 
$(A-\lambda I)v=0$, where $v$ is corresponding eigenvector.
If (by luck)you have $rank(A-\lambda I)=2$ exactly, then this particular method can be used which uses just the first $2$ rows of $(A-\lambda I)v=0$ to solve using cross-multiplication method. An insight is as follows:
$ax+by+cz=0$
$px+qy+rz=0$
Then $\frac{x}{br-cq}=\frac{-y}{ar-pc}=\frac{z}{aq-bp}$
