use either outward or inward arrows to define a "positive torque" in accordance with the right-handed rule,

draw graphs showing the variation of torque and twist along a shaft

use superposition to combine the torques and twists due to external torques applied at two different locations (B, C).

Remember

the RH rule only defines what a torque vector (arrow) means; as with any mechanics problem, it is up to you to draw a diagram with arrows showing what you have chosen as the positive direction for each force or torque vector.

"arrows outward": positive torque makes the shaft twist such that, looking along it, the angle of twist is increasingly clockwise as one looks further away. This torque would unscrew a conventional nut and bolt.

"arrows inwards": positive torque makes the shaft twist such that, looking along it, the angle of twist is increasingly anti-clockwise as one looks further away. This torque would tighten up a nut and bolt.

Suggestions:

Set the slider for torque C = 0. Observe how the torque B splits between the shafts to left and right of the gear wheel at B:
With only the left wall holding the shaft from rotating, all the torque from B is taken in the left-hand shaft. This shaft twists. The other shafts carry no torque so the angle of rotation is constant all the way along them.

Change the definition of torque vector T from outwards to inwards. Satisfy yourself that the graph shows the expected (positive or negative?) value for torque.
Similarly with only the right wall holding the shaft from rotating, all the torque from B is taken in the right-hand shaft. Again, satisfy yourself about the positive/negative sense in terms of the T definition.

With both walls holding the shaft ends, the torque from B splits to left and right.

The three shaft sections are of equal length so the single (left) section has twice the stiffness GJ/L of the double-length section to the right of B. The torque therefore splits to left and right in the ratio 2:1 i.e. 2/3 left, 1/3 right but we must remember that the sign convention gives us torque of opposite sign each side of the gear i.e. 2B/3, -B/3 or -2B/3, B/3 depending on our choice of "positive". [Remember that a shaft with torque T applied at each end is in equilibrium! With end torques T1 and T2, the external torque would be T2-T1 (arrows outwards).

Now vary torque C. The torque graph and the twist graph now show the superimposed effects of both the B and C torques.
In particular, try the double-wall case with:
(a) B = C. The systems is symmetrical and there is on torque in the centre section.
(b) B = -C. The system is anti-symmetrical and the angle of twist at the centre-point is zero.