We can assign variables to numbers of people involved in one or more projects:
A ss only
B pi only
C pd only
D ss+pi
E ss+pd
F pi+pd
G all three projects
We know that each person is involved in at least one project, so:
A+B+C+D+E+F+G=60
A+D+E=23 (ss) }
B+D+F=28 (pi) } excludes those working on all three*
C+E+F=26 (pd) }
E=7, F=12, G=6
A+D=23-7=16=28-12=B+D so B=A, D=16-A
C=26-(7+12)=7
2A+7+(16-A)+7+12+6=60,
A+7+16+7+12+6=60,
A=60-48=12 and D=16-A=16-12=4.
So (A,B,C,D,E,F,G)=(12,12,7,4,7,12,6), therefore 12 work on each of ss and pi and 7 work on pd.
* If the figures were to include those involved on all 3 projects, we get a negative result for one group of people, so there would be no sensible solution. The solution to this question depends on the interpretation of the figures provided.