The three numbers are 12, 18 and 24
Arithmetic progression
Let the 3 number in arithmetic progression be:
a-d, d, a+d ...
If their sum is 3, then;
a-d+d+a+d = 3
2a + d = 3 ........... 1
If the sum of their squares is 11, then;
(a-d)² + d² + (a+d)² = 11
a²-2ad+d²+d²+a²+2ad+d² 11
2a²+3d² = 11 ....... 2
Solving the equations simultaneously, d = 6 and a = 12
First-term = 12
second term = 18
Thirs term = 24
Hence the three numbers are 12, 18 and 24
Hope this helps you!!!!!! :D
I Got 0.01041666666 As An Answer <span />
Keep the bases of 6 and add exponents: 
Keep the bases of 11 and add exponents: 
So its C.
hope that helps :)
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