Given,
x+y+z=12 (x+y+z)2=x2+y2+z2+2(xy+yz+zx)
⇒144=50+2(xy+yz+zx) ⇒xy+yz+zx=×94=47 Now,
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
⇒216−3xyz=12[50−47]⇒3xyz=216−36⇒xyz=60 There are only 3 digits i.e.
3,4,5 whose sum is 12 , sum of square is 50 , sum of cube is 216 and product is 60.
Number formed by three digits is
3!=6 i.e.
(3,4,5),(3,5,4),(4,3,5),(4,5,3),(5,3,4). (5,4,3).