$\vec{a}=(2, -4, 3), \vec{b}=(-4, 8, -6)$ are elements of $\Bbb R^3$. Notice that $\vec b = -2\vec a$, thus the two vectors are collinear. So the space of vectors that are orthogonal the both of these vectors will just be the space of vectors orthogonal to the line that passes through both of them. Can you see that this space will be a plane?
So you just need to specify a plane with a vector equation. The vector equation of a plane is $\vec r(s,t) = \vec us + \vec vt + \vec c$.
We just need to find any two non-collinear vectors orthogonal to $\vec a$ or $\vec b$ (any we find orthogonal to one will automatically be orthogonal to the other).
So we need $(x,y,z) \cdot (2,-4,3) = 2x-4y+3z=0$. Being a linear equation in three variables, we should just be able to choose two of the variables and solve for the last (though don't choose them both zero or you'll just end up with the zero vector which is collinear with every other vector). Let's choose $x=3$ and $y=0$. Plugging in, we see that $z=-2$. So one vector orthogonal to $(2,-4,3)$ is $(3,0,-2)$. Now let's choose $x=0$ and $y=3$ (You could choose the the two numbers to be whatever you like, but notice I chose them so that I'd get integer solutions because no one likes unnecessary fractions). Then $z=4$. So another vector orthogonal to $(2,-4,3)$ is $(0,3,4)$. Notice that $(3,0,-2)$ and $(0,3,4)$ are not collinear (they are not scalar multiples of each other).
So let $\vec u=(3,0,-2)$ and $\vec v=(0,3,4)$. Then to find $\vec c$ we need any point on the line $\operatorname{span}(2,-4,3)$. $(0,0,0)$ is in that span. So let $\vec c = \vec 0$.
Then the equation representing our plane -- and thus every vector orthogonal to $\vec a$ and $\vec b$ -- is just $$\vec r(s,t) = (3,0,-2)s + (0,3,4)t$$
Yes, this is always possible. Choose an arbitrary vector $\vec{w}$ that lies in $V \setminus \text{span}(\vec{a}_1,\dots,\vec{a}_m)$. The orthogonal projection of $\vec{w}$ on $\vec{a}_1$ is
\begin{equation}
\vec{w}_1=\frac{\vec{a}_1 \cdot \vec{w}}{\vec{a}_1 \cdot \vec{a}_1} \vec{a}_1 = \frac{\vec{a}_1 \cdot \vec{w}}{\left| \vec{a}_1 \right|^2} \vec{a}_1
\end{equation}
You can check that the vector $\vec{w}-\vec{w}_1$ is orthogonal to $\vec{a}_1$.
In the same manner, project $\vec{w}-\vec{w}_1$ orthogonally on $\vec{a}_2$:
\begin{equation}
\vec{w}_2=\frac{\vec{a}_2 \cdot (\vec{w}-\vec{w_1})}{\left| \vec{a}_2 \right|^2} \vec{a}_2
\end{equation}
You can again check that the vector $(\vec{w}-\vec{w}_1)-\vec{w}_2$ is orthogonal to $\vec{a}_2$. However, there is no garantee that $(\vec{w}-\vec{w}_1)-\vec{w}_2$ is orthogonal to $\vec{a}_1$, except if the given vectors $\vec{a}_1,\dots,\vec{a}_m$ are already orthogonalised (which can be done by the Gram-Schmidt process).
Repeat tis strategy: calculate $\vec{w}_3,\vec{w}_4,\dots,\vec{w}_m$ in the same way. The vector you were looking for is $\vec{w}-\vec{w}_1-\vec{w}_2-\dots-\vec{w}_m$.
I hope this answer is helpful for you.
Best Answer
Maybe this will help: How about finding a basis containing that vector, extending the basis to a basis for the space, and then applying Gram-Schmidt to get an orthonormal basis?
One way to do this is by using the fundamental theorem of linear algebra: http://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra . Using the 4d-vector as the first row of a $4 \times 4$ matrix $M$ , then finding the nullspace of that matrix, since ker$M$ =orthogonal complement of the row space. This will extend the 4d-vector into a basis for $\mathbb R^4$
Then you can apply Gram-Schmidt algorithm will give you three vectors orthogonal to the original vector.