A circle is inscribed in the triangle ABC whose sides are given as AB = 10, BC = 8, CA= 12 units as shown in the figure. The value of AD$$\times$$BF is:

Given,

AB = 10, BC = 8, CA = 12

AD and AE are tangents to the circle from point A

$$=$$>  AD = AE

Let AD = AE = a

BE and BF are tangents to the circle from point B

$$=$$>  BE = BF

Let BE = BF = b

CD and CF are tangents to the circle from point C

$$=$$>  CD = CF

Let CD = CF = c

AB = 10

$$=$$>  AE + BE = 10

$$=$$>  a + b = 10 .................(1)

BC = 8

$$=$$>  BF + CF = 8

$$=$$>  b + c = 8 ...................(2)

CA = 12

$$=$$>  CD + AD = 12

$$=$$>  c + a = 12 ..................(3)

Solving (1) - (2)

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

Soving (3) + (4)

2a = 14

$$=$$>  a = 7

From (1), 7 + b = 10

$$=$$>  b = 3

From (3), c + 7 = 12

$$=$$>  c = 5

$$\therefore\ $$AD$$\times$$BF = a$$\times$$b = 7$$\times$$3 = 21 units

Hence, the correct answer is Option A