1. Count total points

There are:

• 1 point at the origin $O$
• 12 points $P_1, P_2, \dots, P_{12}$ on line $L_1$
• 9 points $Q_1, Q_2, \dots, Q_9$ on line $L_2$

So total number of points is $1+12+9=22.$

If no three points were collinear, the number of triangles would be $\binom{22}{3} = \frac{22\cdot 21\cdot 20}{6}=1540.$

2. Subtract collinear triples

A triangle cannot be formed if all 3 chosen points are collinear.

The only collinear sets of 3 or more points lie on the two given lines.

On line $L_1$

Points on $L_1$ are: $O, P_1, P_2, \dots, P_{12}$ So there are $13$ points on $L_1$.

Number of ways to choose 3 collinear points from these is $\binom{13}{3} = \frac{13\cdot 12\cdot 11}{6}=286.$

On line $L_2$

Points on $L_2$ are: $O, Q_1, Q_2, \dots, Q_9$ So there are $10$ points on $L_2$.

Number of ways to choose 3 collinear points from these is $\binom{10}{3} = \frac{10\cdot 9\cdot 8}{6}=120.$

3. Compute number of triangles

Hence total number of triangles is $\binom{22}{3}-\binom{13}{3}-\binom{10}{3}$ $=1540-286-120=1134.$

4. Check options

$1134$ corresponds to Option C.

Therefore, the correct answer is $\boxed{1134}.$