Step 1: Express $m$ using the linear relation.
From $2l - m + 3n = 0$, we get $m = 2l + 3n$. This allows us to eliminate $m$ from the quadratic relation.
Step 2: Substitute into the quadratic relation.
Substituting $m = 2l+3n$ into $l^2 + mn - 6n^2 = 0$: \[l^2 + (2l+3n)n - 6n^2 = 0\] \[l^2 + 2ln + 3n^2 - 6n^2 = 0\] \[l^2 + 2ln - 3n^2 = 0\]
Step 3: Factorise to find the two cases.
Treating as a quadratic in $l$: \[(l + 3n)(l - n) = 0\] So either $l = n$ or $l = -3n$.
Step 4: Find direction cosines for the first line ($l = n$).
When $l = n$: $m = 2n + 3n = 5n$. Direction ratios are $(l:m:n) = (1:5:1)$. Magnitude $= \sqrt{1+25+1} = \sqrt{27} = 3\sqrt{3}$. Direction cosines: \[(l_1, m_1, n_1) = \left(\frac{1}{3\sqrt{3}},\ \frac{5}{3\sqrt{3}},\ \frac{1}{3\sqrt{3}}\right)\]
Step 5: Find direction cosines for the second line ($l = -3n$).
When $l = -3n$: $m = 2(-3n)+3n = -3n$. Direction ratios are $(-3:-3:1)$. Magnitude $= \sqrt{9+9+1} = \sqrt{19}$. Direction cosines: \[(l_2, m_2, n_2) = \left(\frac{-3}{\sqrt{19}},\ \frac{-3}{\sqrt{19}},\ \frac{1}{\sqrt{19}}\right)\]
Step 6: Compute $|l_1 l_2| + |m_1 m_2|$.
\[|l_1 l_2| = \frac{1}{3\sqrt{3}} \cdot \frac{3}{\sqrt{19}} = \frac{1}{\sqrt{3}\cdot\sqrt{19}} = \frac{1}{\sqrt{57}}\] \[|m_1 m_2| = \frac{5}{3\sqrt{3}} \cdot \frac{3}{\sqrt{19}} = \frac{5}{\sqrt{57}}\] Sum $= \dfrac{6}{\sqrt{57}} = \dfrac{6}{\sqrt{57}} \cdot \dfrac{\sqrt{57}}{\sqrt{57}} = \dfrac{6\sqrt{57}}{57} = \dfrac{2\sqrt{57}}{19}$. Checking option 2: $\dfrac{2\sqrt{3}}{\sqrt{19}} = \dfrac{2\sqrt{3}\cdot\sqrt{19}}{19} = \dfrac{2\sqrt{57}}{19}$. They are equal.
Step 7: State the final answer.
\[ \boxed{\frac{2\sqrt{3}}{\sqrt{19}}} \]