As can be seen on Julius' drawing above, the components of current phasors IRY and -IBR that are in-phase with each other add together to form phasor IR, which represents the line current at terminal R.
The components of IRY and -IBR that are in quadrature (i.e., at right angles) to the output phasor cancel out and do not appear in the line current. Instead, these components represent the current that flows between the loads on B-R and R-Y through their common connection at terminal R. This behavior is analogous to the current that flows between balanced L1-N and L2-N loads on a split-phase system without contributing a current on the neutral conductor.
In the case presented by the OP the current that flows between two resistive loads at their common connection in the delta configuration is sin(30°) x 50A = 0.5 x 50A = 25A. This current can be measured by arranging the wires connecting the delta to go through a clamp meter as in the drawing below, where the clamp meter is indicated by oval #2. In this case the indicated current should be 50A because two wires which each flow 25A in the same direction are encircled by the clamp's sensor.
Such a measurement would be informative about where the "missing" current is going when you use a 1.73x factor instead of a 2x factor when load currents are summed to obtain the line current in a delta. Otherwise, you probably wouldn't have a need to do this particular measurement.
The line current measured at location #1 would be 2 x cos(30°) x 50A =2 x 0.866 x 50A = 86.6A as mentioned in the posts above.