Let

$x$ = the hundreds digit

$y$ = the tens digit

$z$ = the units digit

$100x + 10y + z$ → the original number

$100x + 10z + y$ → the tens and units digits are interchanged

$100y + 10x + z$ → the hundreds and tens digits are interchanged

The sum of the digits of a three-place number is 19

$x + y + z = 19$ → Equation (1)

If the tens and units digits are interchanged the number is diminished by 27

$(100x + 10z + y) = (100x + 10y + z) - 27$

$-9y + 9z = -27$

$-y + z = -3$

$z = y - 3$ → Equation (2)

If the hundreds and tens digits are interchanged the number is increased by 180

$(100y + 10x + z) = (100x + 10y + z) + 180$

$90y - 90x = 180$

$y - x = 2$

$y = x + 2$ → Equation (3)

Substitute y = x + 2 to Equation (2)

$z = (x + 2) - 3$

$z = x - 1$

Substitute y = x + 2 and z = x - 1 to Equation (1)

$x + (x + 2) + (x - 1) = 19$

$3x = 18$

$x = 6$

From Equation (3)

$y = 6 + 2$

$y = 8$

From Equation (2)

$z = 8 - 3$

$z = 5$

The number is

$100x + 10y + z = 685$ *answer*