a.) (3,3) (3,9) (9,3)
b.) (0,0) (0,6) (6,0)
c.) (3,3) (0,9) (9,0)
d.) (0,0) (0,9) (9,0)

Explanation:

The original three vertices are A = (0,0), B = (0,3) and C = (3,0). Each point is formed by reading the columns from top to bottom.

Multiply those coordinates by 3

A = (0,0) turns into A ' = (0,0) ... ie there's no change here

B = (0,3) turns into B ' = (0, 9)

C = (3,0) turns into C ' = (9, 0)

The three points A', B', and C' are three times further away from the origin compared to the original points A, B, C.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!