Solution:

Given, AD : DB = 1 : 3, BE : EC = 3 : 2 and CF : FA = 4 : 1.

Let us assume that the triangle ABC has an area of 'x'.

Area of triangle ABC = x

We know that area of triangle = ½ × base × height.

Let us find the height of triangle ABC from point A.

Let h be the height of triangle ABC from point A.

Therefore, area of triangle ABC = ½ × AB × h = ½ × AC × h

⇒ AB = AC

Let us find the length of AD.

Let AD = x

Therefore, DB = 3x

Similarly, let BE = 3y and EC = 2y.

Let CF = 4z and FA = z.

From triangle ADF, we can find the height of triangle DEF from point D.

Let h1 be the height of triangle DEF from point D.

Therefore, area of triangle DEF = ½ × DF × h1

From triangle BDE, we can find the height of triangle DEF from point E.

Let h2 be the height of triangle DEF from point E.

Therefore, area of triangle DEF = ½ × DE × h2

From triangle CDF, we can find the height of triangle DEF from point F.

Let h3 be the height of triangle DEF from point F.

Therefore, area of triangle DEF = ½ × DF × h3

Adding the above three equations, we get

Area of triangle DEF = ½ × (DF × h1 + DE × h2 + DF × h3)

We know that DF = AD + AF = x + z and DE = BE – BD = 3y – 3x

Substituting the values of DF and DE in the above equation, we get

Area of triangle DEF = ½ × [(x + z) × h1 + (3y – 3x) × h2 + (x + z) × h3]

Let us find the ratio of area of triangle DEF to that of triangle ABC.

Area of triangle DEF/Area of triangle ABC = [½ × [(x + z) × h1 + (3y – 3x) × h2 + (x + z) × h3]]/x

= ½ × [(x + z) × h1 + (3y – 3x) × h2 + (x + z) × h3]/x

= ½ × [(x + z) × h1/x + (3y – 3x) × h2/x + (x + z) × h3/x]

= ½ × [(AD/AC + FA/AC) × h1 + (BE/BC – AD/AC) × h2 + (FA/AC + CF/BC) × h3]

Substituting the given values of AD : DB, BE : EC and CF : FA, we get

Area of triangle DEF/Area of triangle ABC = ½ × [(1/4 + 1/5) × h1 + (3/5 – 1/4) × h2 + (1/5 + 4/5) × h3]

= ½ × [(9/20) × h1 + (7/20) × h2 + h3]

= (9h

9:50