Take number as x and y

so number is 10x + y [x as tens digit and y as unit digit]

the sum of this digit with it's reverse is 143 so,

new reverse no is 10y + x

10x + y + 10y + x = 143

11x + 11y =143

x + y = 13............[1]

now sum of squares of digits so x^2 + y^2

number is 3 less than the sum of the squares of its digits.

so , [x^2 + y^2 ] - 3..............[2]

the number and euation 2 is same so

10x + y = [x^2 + y^2 ] - 3

10x + y = [169 - 2xy] - 3

10x + y + 2xy = 166 ...... [3]

now [3]- 10*[1]

- 9y +2xy = 36

y[ 2x-9] =36 put value of y from 1

13-x [2x-9] =36

2x^2 - 35x + 153 = 0

solving this by √D = b²-4ac [a = 2, b = -35, c = 153]

=1225-1224

=1

√D =1

now x= -b+√D÷2a or x=-b-√D÷2a

x=9 and x=34/4

so we take x = 9

so y = 13 - 9 = 4

number is 10x + y = 10*9 + 4 =** 94**