Another way to find the zero shear point is to build a running total of the vertical forces from the left support instead of jumping straight to the segment values.
First get the reactions using $\Sigma M_P = 0$ and $\Sigma F_y = 0$:
Now scan the beam from P to S and keep a running sum of upward minus downward forces met so far:
Just right of P: running sum $= +13.33$ kN (only $R_P$ counted so far)
Just right of Q: running sum $= 13.33 - 10 = +3.33$ kN (the 10 kN load subtracted)
Just right of R: running sum $= 3.33 - 20 = -16.67$ kN (the 20 kN load subtracted)
Just right of S: running sum $= -16.67 + 16.67 = 0$ kN (the reaction $R_S$ closes the diagram)
The running sum is positive all the way up to R and negative right after R, so the sign change, and with it the zero crossing of the shear force diagram, happens exactly at R. Between P and Q, and between Q and R, the running sum never touches zero because no load acts in between to change it, it only changes value at a load or a reaction point.
So the zero shear point sits at R, which is also where the bending moment will be largest, since the moment is the running area under the shear diagram and it stops growing once the shear changes sign.
