Let their common differences be d1,d2,d3
Given d1,d2,d3 are in H.P.
⇒1d1,1d2,1d3 are in A.P.
S1=112[2+(11−1)d1]=11(1+5d1)
⇒d1=S1−1155
Similarly, d2=S2−1155,d3=S3−1155
⇒55S1−11,55S2−11,55S3−11 are in A.P.
⇒55(1S2−11−1S1−11)=55(1S3−11−1S2−11)
⇒S1−S2S1−11=S2−S3S3−11
⇒11(S1−2S2+S3)=2S3S1−S1S2−S2S3
⇒2S3S1−S1S2−S2S3S1−2S2+S3=11