Lets

x = first number

y = second number

z = third number

We have the following system of equations

y = 2*x - 5 (number is 5 less than 2 times the first)

z = 6*(x/2) (third number is 6 times more than half the first)

x + y + z = 351 (he sum of the numbers is 351)

if we substitute for y and z and the third equation we get

x + 2*x - 5 + 3*x = 351

6*x = 356

x = 356/6 = 178/3 = 59.3333 approximately

y = 2*(178/3) - 5 = (356 - 15)/3 = 341/3 = 113.667 approximately

z = 3*x = 178

The numbers we are looking for are 178/3, 341/3 and 178.