Given the line (1-n)*x + n*y -1 = 0 calculate n for the line is perpendicular to another line that passes through (-2,3) , (1,2).

We  shall determine the line through (-2,3) and
(1,2).

The line through (x1,y1) and (x2,y2) is given
by:

y -y1 = {(y2-y1)/(x2-x1)}
(x-x1)...(1)

(x1 ,y1) = (-2,3) and (x2, y2) = 1,2).
Substituting in the formula (1), we get:

(y-3) = {(2-3)/(1-
-2)}(x --2)

y -3 = (-1/3)
(x+2)

(3y-9) = (-(x+2)

 x+3y
-9+2 = 0

x+3y -7 = 0.This is the line through the given 2
points.

Any line perpendicular to this is  got by reversing
the coefficients of x and y and putting a minus sign to one of the
coefficints.

So x+3y -7 = 0 has the perpendicular is of the
form:

3x-y +k = 0. Comparing this line with the line (1-n)+
ny -1 = 0. If both equations  are of the same line then the coefficits of x , y and
contant terms should have the same proportion:

(1-n)/3 =
n/-1 = k/-1.

From the first two equations, (1-n)/3 =
n/(-1)

n-1 = 3n

-1 =
3n-n

-1= 2n

n =
-1/2.

So n = -1/2.