设百位为x,十位为y,个位为z
则有：x+y+z = 11 (1)
100z + 10y + x -(100x + 10y +z) = 693
==> 99z - 99x = 693
==> z - x = 7
==> x = z-7 (2)
100x + 10z + y -(100x +10y + z) = 54
==> 9z - 9y = 54
==> z - y = 6
==> y = z-6 (3)
将(2)和(3)代入(1)中得
z-7 + z-6 + z = 11
==> 3z = 24
==> z = 8
代入(2)中解得 x = z-7 = 8-7 = 1
代入(3)中解得 y = z-6 = 8-6 = 2
所以这个三位数是 128