In a triangle ABC, length of the side AC is 4 cm more than 2 times the length of the side AB. Length of the side BC is 4 cm less than the three times the length of the side AB. If the perimeter of $$\triangle$$ABC is 60 cm, then its area (in cm$$^2$$) is:

Length of the side AC is 4 cm more than 2 times the length of the side AB.

b = 2c + 4......(1)

Length of the side BC is 4 cm less than the three times the length of the side AB.

a = 3c - 4.......(2)

Perimeter of $$\triangle$$ABC is 60 cm.

a + b + c = 60

3c - 4 + 2c + 4 + c = 60

6c = 60

c = 10 cm

b = 2c + 4 = 24 cm

a = 3c - 4 = 26 cm

Half of the perimeter(s) = 30 cm

Area of the triangle = $$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$$

= $$\sqrt{30\left(30-26\right)\left(30-24\right)\left(30-10\right)}$$

= $$\sqrt{30\left(4\right)\left(6\right)\left(20\right)}$$

= $$\sqrt{120\times120}$$

= 120 cm$$^2$$

Hence, the correct answer is Option D