Adding one won't work here!

Take the first number as $a$

The next ones are $a+1=b$ and $a+2=c$

$a^{2} + (a+1)^{2} + (a+2)^{2} = 110$

$a^{2} + a^{2} + 1 + 2a + a^{2} + 4 + 4a = 110$

$3a^{2} + 6a + 5 = 110$

$3a^{2} + 6a + 5 -110 = 0$

$3a^{2} + 6a - 105 = 0$

Taking 3 as common in all.

$3 x (a^{2} + 2a - 35) = 3 x 0$

$a^{2} + 2a - 35 = 0$

Factorizing it; we get

$a-5$ $a+7$ = $0$

$a = 5$

$a = -7$

As it says positive; 5 must be the first number. Therefore; $5+1 = 6$ and $5+2 = 7$

$5+6 + 7 = 18$

take 3 consecutive numbers x,x+1andx+2. put in given condition which after simplification will lead to a quadratic equation with roots 5 &-7.. since required number being positive 5 is the only choice so required consecutive numbers are 5,6&7 having sum 18...