A (2, 3+a) , B(a-1, 2a) C (3, 5)

The point are on the same line,

Then slope AB = slope BC = slope (AC

slope AC = ((5-(3+a)/(3-2)= (2-a)

slope BC= (5-2a)/(3-(a-1) = (5-2a)/ (4-a)

We know that slope AC = slope BC

==> 2-a = (5-2a)/(4-a)

Cross multiply:

==> (2-a)(4-a) = 5-2a

==> 8 -6a +a^2 = 5-2a

Group similar terms:

==> a^2 -4a +3 = 0

==> (a-3)(a-1) = 0

==> There are no solutions:

a1= 3

a2= 1

The points are A(2,3+a), B(a-1 , 2a), C(3,5)

We know that slope of a line MN = (yN- yM)/(xM-xN)

If the points are A< B and C are collinear, then

Slope of AC = slope of AB. Or

(5-(3+a))/(3-2 ) = (2a -(3+a)/(a-1-2)

(2-a) = (a-3)/(a-3)

2-a = 1

2-1 = a. Therefore a =1.

If the given points belong to the same line, the determinant formed by these points has to have zero value.

Let's build the determinant:

2 (3+a) 1

(a-1) 2a 1

3 5 1

We'll calculate the determinant using triangle method:

det = 2*2a*1 + 5*(a-1) + 3*(3+a) - 3*2a - 5*1*2 - (a-1)*(3+a)

det = 4a + 5a - 5 + 9 + 3a - 6a - 10 - 3a - a^2 + 3 + a

We'll combine like terms:

4a - a^2 - 3 = 0

We'll multiply by -1:

a^2 - 4a + 3 = 0

We'll apply the quadratic formula:

a1 = [4 + sqrt(16 - 12)]/2

a1 = (4+2)/2

**a1 = 3**

a2 = (4-2)/2

**a2 = 1**