The sum of the squares of three consecutive natural number is 149.

Solution:

Let three consecutive integer be $x,(x+1)$ and $(x+2)$

Then according to question

$x^{2}+(x+1)^{2}+(x+2)^{2}=149$

$x^{2}+x^{2}+2 x+1+x^{2}+4 x+4=149$

$3 x^{2}+6 x+5-149=0$

$3 x^{2}+6 x-144=0$

$3 x^{2}+6 x-144=0$

$3\left(x^{2}+2 x-48\right)=0$

$x^{2}+2 x-48=0$

$x^{2}+8 x-6 x-48=0$

$x(x+8)-6(x+8)=0$

$(x+8)(x-6)=0$

$(x+8)=0$

$x=-8$

Or

$(x-6)=0$

$x=6$

Since, x being a positive number, so x cannot be negative.

Therefore,

When $x=6$ then other positive integer

$x+1=6+1$

$=7$

And

$x+2=6+2$

$=8$

Thus, three consecutive positive integer be $6,7,8$