Linear Algebra - Homogenous Equations Find a homogenous system of linear equations in five unknowns whose solution space consists of all vectors $\Re^5$ that are orthogonal to the vectors: 

$\mathbf v_1$$=\langle3,0,1,2,3\rangle$;

$\mathbf v_2$$=\langle-3,1,0,-1/2,-1\rangle$;

$\mathbf v_3$$=\langle6,0,1,2,-3\rangle$. 

What kind of geometric object is the solution set? Find a general solution of the system and confirm that the solution space has the orthogonality and the geometric properties that were stated.

I was told this problem would be really good practice for linear algebra however I have no idea how to approach this problem and would appreciate any guidance.
We are looking for all $\mathbf x=\langle x,y,z,u,v\rangle \in \Re^5$ such that 

$\mathbf x \cdot \mathbf v_j=0$ for $j=1,2,3$, where $ \mathbf x \cdot \mathbf v_j$ denotes the usual inner product on $\Re^5$.
A vector $(x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5$ is orthogonal to a vector $(a_1, a_2, a_3, a_4, a_5) \in \mathbb{R}^5$ if and only if $a_1 x_1 + a_2 x_2 + a_3 x_3 + a_4 x_4 + a_5 x_5 = 0$. Here, the set of vectors you are looking for is the set of solutions of the system :
\begin{equation}
  \left\{
      \begin{aligned}
        3x_1&amp; &amp; +x_3&amp; +2x_4&amp;+3x_5&amp;=&amp;0\\
        -3x_1&amp;+x_2&amp; &amp;-\frac{1}{2}x_4&amp;-x_5&amp;=&amp;0\\
        6x_1&amp; &amp;+x_3&amp;+2x_4&amp;-3x_5&amp;=&amp;0\\
      \end{aligned}
    \right.
\end{equation}

Now you get a linear system and you can resolve it with your favorite method.