answer is  65292

let 5 digits be

a b c d e

now,

eq1=>    d= 4+b,
eq2=>    c = b- 3
eq3=>     a =3e,

As a,b,c,d,e are single digits can varies from 0-9

so
bmax <= 5 , from eq1
bmin >= 3, from eq

similarly
dmax = 9
dmin = 7

cmax = 2
cmin= 0

e can be     0, 1,2,3
a will be     0,3,6,9 respectively

So ranges are
0<=a<=9
3<=b<=5
0<=c<=2
7<=d<=9
0<=e<=3

Possible pairs are
ab, ac, ad, ae,     bc, bd, be,     cd, ce,      de

Also there are only three pairs whose sum should be 11
there are 9 cases lets find one by one
a.) if b= 3 => then  c = 0, d= 7
(i)  a = 9,  e= 3,
or  (ii) a = 6,  e = 2,
or (iii) a = 3, e =1

b.) if b = 4 => then c=1, d=8
(i)  a = 9,  e= 3,
or  (ii) a = 6,  e = 2,
or (iii) a = 3, e =1

c.) if b = 5 => then c=2, d= 9
(i)  a = 9,  e= 3,
or  (ii) a = 6,  e = 2,
or (iii) a = 3, e =1

So from this only c(ii) shows the desired result
a+b= 11, c+d =11, d+e = 11

so answer is
a=6,b=5,c=2,d=9,e=2

ie. 65292

answer is 65292. Explanation: Since the 4th digit is 4 greater than the second digit, and the 3rd digit is 3 less than the 2nd digit  we have only 2 options for the 2nd-4th digits: 529 and 418. To get 11 from 2 digits it can be 9+2, 8+3 , 7+4 , 6+5.  If we look on the 2nd option 418 we can see that that if (without taking in consideration the fact that the 1st digit is thrice the last digit) choose the combination of 34183 or 74183 we will get only 2 combinations of 11. So 418 in the middle is not possible. So we are left with 529. So since the 1st digit is thrice the last digit it gives us 3,1 or 6,2 or 9,3. The only possible choice is 6,2 because the other 2 options will not give us that there are 3 pairs whose sum is 11. (35291 will give one pair and 95293 will give 2 pairs)