If sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2, then ∑k=12 x100k+y106k∑x207y207 is

We have, sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2.It is possible only whensin−1⁡x=π2,sin−1⁡y=π2 and sin−1⁡z=π2⇒x=1,y=1 and z=1∴ ∑k=12 x100k+y106k∑x207y207=x100+y106+x200+y212x207y207+y207z207+z207x207=1+1+1+11+1+1=43