sum of squares of three consecutive positive even integer is 200,while their mean is 2(x+1).
Let first positive even integer=x
So second consecutive even integer= x+2
Third consecutive even integer= x+4
x2+(x+2)2+(x+4)2=200
x2+x2+4x+4+x2+8x+16=200
3x2+12x+20=200
3x2+12x-180=0
x2+4x-60=0
x2+10x-6x-60=0
x(x+10)-6(x+10)=0
(x-6)(x+10)=0
x=6, x=-10
x= -10 cannot be the value of x since the integers are positive
so the three consecutive even integers are 6,8 & 10