Find abc?

It is given that a,b and c are three consecutive real numbers .

Therefore, let a=n , b=n+1 , c=n+2

a^2+b^2+c^2 = 974 or n^2+ (n+1)^2 +(n+2)^2 =974 or 3n^2+6n+5=974 . On solving we get n=17 (it has to a positive real number as mentioned)

a=n=17 , b=n+1=18 , c=n+2=19

Therefore abc= 17x18x19 = 5814

$a=b-1,c=b+1$ ,the equation becomes $(b-1)^2+b^2+(b+1)^2=974,3b^2+2=972,b^2=324,b=18,abc=17\times18\times19=5814$