[The lady above showed that your problem has no solution as
you stated it.  That's because you did not give the sum of 
the numbers and it appeared that the sum of the numbers was 
also 126.  But since that gives no solution, I arbitrarily 
assumed it was 18 since that or -18 were the only numbers 
that would give a reasonable answer.  But please be careful 
next time to copy the problem exactly.]

Let the three numbers be a-d, a, and a+d.

Those are in arithmetic progression because:

1. If we add d to the first term, which is a-d, we get 
(a-d)+d or a, which is the second term.
2. If we add d to the second term, which is a, we get
a+d which is the third term.

1)     (a-d)+a+(a+d) = 18
2)     (a-d)²+a²+(a+d)² = 126

Simplifying equation 1):

3a = 18
 a = 6

Substituting a = 6 in the equation 2):

(6-d)²+6²+(6+d)² = 126

Simplifying:

(36-12d+d²)+36+(36+12d+d²) = 126
    36-12d+d²+36+36+12d+d² = 126 
                   2d²+108 = 126

Divide through by 2
                   
                     d²+54 = 63
                        d² = 9
                         d = ±3
If d = +3, the three terms are, since a = 6

a-d = 6-3 = 3  
a = 6 
a+d = 6+3 = 9

If d = -3, the three terms are, since a = 6

a-d = 6-(-3) = 6+3 = 9  
a = 6 
a+d = 6+(-3) = 6-3 = 3

So regardless of whether the the numbers are 

3,6,9 or 9,6,3,

The smallest is 3.

Edwin