`A={x:x in R and x^(@)-8x+12=0}`
`:. x^(2)-8x+12=0 or x^(2)-(2+6)x+12=0`
or `x^(2)-2x-6x +12 =0`
or `x(x-2) - 6(x-2)=0`
or `(x-2)(x-6)=0 rArr x=2,6`
`:. A={2,6} and B={2,4,6},C={2,4,6,8,...},D={6}`
`:. 6` is an element in all sets and it is one elements of singleton set D
`:.D subA,D subB,D subC`
`because` In `A = {2,6} and B = {2,4,6}`, each element of A is in B
`A sub B`
Similarly, in `B = {2,4,6} and C = {2,4,6,8,...}` each element of B and A is in C
`A sub C and B sub C`
Therefore, `D subB,B subC,A subC,D subA,D subC, A subB`.