First, compute angle $\theta := \angle (a,b)$ between $a$ and $b$ using cosine law.
$$
\left\langle a, b \right\rangle = \left\| a\right\| \left\| b\right\| \cos \theta
\implies
\cos \theta = \frac{\left\langle a, b \right\rangle}{\left\| a\right\| \left\| b\right\| },
$$
where $\left\langle \,\cdot\,, \cdot \,\right\rangle$ is inner product of vectors $a$ and $b$.
Note that
$$
\left\| a \right\| = \sqrt{\left\langle a, a \right\rangle }
= \sqrt{\left\langle 2m - 2n , 2m - 2n \right\rangle }
= 2 \sqrt{\left\langle m - n , m - n \right\rangle }
= 2 \sqrt{ \left\langle m,m \right\rangle - 2 \left\langle m, n \right\rangle + \left\langle m,m \right\rangle }
= 2\sqrt{\left\|m\right\|^2 - 2 \left\langle m, n \right\rangle + \left\|n\right\|^2}
$$
Since $\left\langle m, n \right\rangle = \left\|m\right\| \left\|n\right\|\cos\angle\left(m,n \right) = 1\cdot 1\cdot \cos \frac{2\pi}{3} = -\frac{1}{2}$,
we have
$$
\begin{aligned}
\left\| a \right\|
&= 2\sqrt{\left\|m\right\|^2 - 2 \left\langle m, n \right\rangle + \left\|n\right\|^2}
= 2\sqrt{1^2 + \frac{2}{2} + 1^2}
= 2\sqrt{3}
\\
\left\| b \right\|
&= \sqrt{\left\langle 3m+6n,3m+6n\right\rangle}
=3\sqrt{\left\|m\right\|^2 + 4\left\langle m,n\right\rangle + 4\left\|n\right\|^2}
=3\sqrt{1 - \frac{4}{2} + 4 }
= 3\sqrt{3}
\\
\left\langle a,b\right\rangle
&= \left\langle 2m-2n,3m+6n\right\rangle
= 6 \left\| m\right\| ^2 + 6 \left\langle m,n\right\rangle - 12 \left\| n\right\|^2
= 6 - \frac{6}{2} - 12 = -9
\end{aligned}
$$
And so
$$
\cos \theta
= \frac{\left\langle a, b \right\rangle}{\left\| a\right\| \left\| b\right\| }
= \frac{-9}{2\sqrt{3} \cdot 3\sqrt{3}}
= -\frac{1}{2}
$$
You are looking for a vector $v$ such that $\angle\left(a,v\right) = \theta/2 $ and $\left\| v\right\| = 10\sqrt 3$.
Assume $x,y$ are coefficients of decomposition of $v$ in terms of $m$ and $n$, i.e. $v = x m + y n$, then we write
$$
v = x m + yn, \quad
\left\| v\right\| = 10\sqrt 3, \quad
\cos \frac{\theta}{2} = \sqrt{\frac{1+ \cos \theta}{2}}
= \sqrt{\frac{1-\frac{1}{2}}{2}}
= \pm \frac{1}{2}
$$
Let us write out what we know about $v$:
$$
\begin{aligned}
\left\langle v,v\right\rangle
& = \left\| v \right\|^2 = 300
\\
\left\langle v,a\right\rangle
& = \left\| v \right\| \left\| a \right\| \cos \frac{\theta}{2}
= 10\sqrt{3}\cdot 2\sqrt{3}\cdot \frac{\pm 1}{2}= \pm 30
\\
\left\langle v,b\right\rangle
& = \left\| v \right\| \left\| b \right\| \cos \frac{\theta}{2}
= 10\sqrt{3}\cdot3\sqrt{3}\cdot\frac{\pm 1}{2}=\pm 45
\end{aligned}
$$
On the other hand,
$$
\begin{aligned}
30 = \left\langle v,a\right\rangle &= \left\langle xm+yn,2m-2n\right\rangle
= 2\left(x \left\| m\right\| ^2 + (y - x) \left\langle m,n\right\rangle - y \left\| n\right\|^2 \right) =
\\ &
= 2\left(x -\frac{y - x}{2} - y \right) = 3x-3y
\\
45 = \left\langle v,b\right\rangle &= \left\langle xm+yn,3m+6n\right\rangle
= 3\left(x \left\| m\right\| ^2 + (y + 2x) \left\langle m,n\right\rangle + 2 y \left\| n\right\|^2 \right)=
\\ &
= 3\left(x -\frac{y + 2 x}{2} + y \right) = \frac{9}{2}y
\end{aligned}
$$
Thus we have a linear system
$$
\begin{cases}
\displaystyle 3x - 3y = 30 \\
\displaystyle \dfrac{9}{2}y = 45
\end{cases}
\implies
\begin{cases}
\displaystyle x = 10 + y\\
\displaystyle y = 10
\end{cases}
\implies
\begin{cases}
\displaystyle x = 20\\
\displaystyle y = 10
\end{cases}
$$
Finally, we write the answer
$$
\bbox[5pt, border: 2pt solid #FF0000]{v = 20 m + 10 n}
$$
One can check that indeed $ \left\| v\right\| = 10\sqrt{3}$.
3D:
Let's say you have two vectors, $\vec{a}$ and $\vec{b}$. They are perpendicular to $\vec{n}$,
$$\vec{n} = \vec{a} \times \vec{b} \tag{1}\label{NA1}$$
If we measure the rotation around the vector $\vec{n}$, the angle between the two vectors is $\varphi$,
$$\cos\varphi = \frac{\vec{a} \cdot \vec{b}}{\left\lVert\vec{a}\right\rVert \, \left\lVert\vec{b}\right\rVert} \tag{2}\label{NA2}$$
In general, we can also utilize $$\sin\varphi = \pm \frac{\left\lVert \vec{a} \times \vec{b} \right\rVert}{\left\lVert\vec{a}\right\rVert\,\left\lVert\vec{b}\right\rVert} \tag{3}\label{NA3}$$
where the sign depends on the orientation (clockwise or counterclockwise) we measure the rotation (around the plane normal $\vec{n}$) from $\vec{a}$ to $\vec{b}$.
If you have a specific unit direction vector $\hat{d}$, either parallel or opposite to $\vec{n}$,
$$\hat{d} = \pm \frac{\vec{n}}{\left\lVert\vec{n}\right\rVert} = \pm \frac{\vec{a}\times\vec{b}}{\left\lVert\vec{a}\times\vec{b}\right\rVert}$$
then you can calculate the angle counterclockwise around $\hat{d}$ using $\eqref{NA2}$ and
$$\sin\varphi = \frac{\hat{d}\cdot(\vec{a}\times\vec{b})}{\left\lVert\vec{a}\right\rVert\,\left\lVert\vec{b}\right\rVert}$$
2D:
In this case, there is only one "side"; the 2D coordinate plane. Here, we can use the 2D analog of the vector cross product:
$$\begin{array}{l}
\vec{a} = ( x_a , y_a ) \\
\vec{b} = ( x_b , y_b ) \\
\vec{a} \cdot \vec{b} = x_a x_b + y_a y_b \\
\vec{a} \times \vec{b} = x_a y_b - x_b y_a \end{array}$$
and
$$\begin{cases}
\cos\varphi = \frac{\vec{a} \cdot \vec{b}}{\left\lVert\vec{a}\right\rVert \, \left\lVert\vec{b} \right\rVert} = \frac{ x_a x_b + y_a y_b}{\sqrt{(x_a^2 + y_a^2)(x_b^2 + y_b^2)}} \\
\sin\varphi = \frac{\vec{a} \times \vec{b}}{\left\lVert\vec{a}\right\rVert \, \left\lVert\vec{b} \right\rVert} = \frac{ x_a y_b - x_b y_a}{\sqrt{(x_a^2 + y_a^2)(x_b^2 + y_b^2)}} \end{cases} \tag{4}\label{NA4}$$
Best Answer
I do not view either $r_1$ or $r_2$ as a "vector." Instead, each of them is a line in a three-dimensional space. In order to describe each line, we have arbitrarily chosen one point on each line (identified by the position vector of that point) and parameterized the rest of the line by a vector parallel to the line.
The choice of point for the position vector really is arbitrary: we could just as well write $$ r_1{:\ } i + 2j - 3k + \xi(i+k) $$ and it would describe exactly the same line as in the question.
We can also substitute a scalar multiple of the direction vector: $$ r_1{:\ } i + 2j - 3k + \phi(2i+2k) $$ is still the same line.
In fact you can get the same line from many different possible position vectors, but you must always use a direction vector that is some multiple of the same direction vector. The different choices of direction vector will always give the same answer from the angle formula (except that there are really usually two angles between two lines, an obtuse angle and an acute angle, and if we multiply one of the directions by a negative number we'll change which of those angles the formula gives us). The different choices of position vector will (except in special cases) give us many different answers.
So if you want the angle between two lines, use the angle between their direction vectors. Intuitively, it is the direction of a line that determines the angle with another line, not the position of the line.