Question 694690
First lets assume the three number are n-1,n,and n+1, since the numbers are consecutive integers. Our equation would be
{{{(n-1)+ 5*n= 3(n+1)+51}}}, which simplifies to 
{{{6*n-1=3*n+54}}}. Now we can solve by subtracting 3n and adding 1 to each side:
{{{3*n=55}}}
that would mean that n is not an integer, and there are no consecutive integers which have those properties. 
Numbers that work out that differ by 1 are {{{52/3}}} {{{55/3}}} {{{58/3}}}