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Explanation:

Applying cosine rule in ∆NPL, we get;
 
Cos60 = 1/2 = [NP2 + NL2 − PL2]/(2 × NP × NL)
 
∴ 1/2 = [(2x)2 + (3x)− PL2]/(2 × 2x × 3x)
 
∴ PL = x√7.
 
Using angle bisector theorem, MN/NL = MP/PL
 
So, MN/3x = MP/x√7
 
∴ MN = MP × (3/√7) ...(I)
 
Applying cosine rule in ∆MNP, we get;
 
Cos60 = 1/2 = [MN2 + NP2 − MP2]/(2 × MN × NP)
 
Let MN = a, so using (I) MP = (√7/3) a or MP2 = (7/9)a2.
 
∴ 1/2 = [a2 + 4x2− (7/9)a2 ]/(2 × a × 2x)
 
∴ a2 − 9ax + 18x2 = 0  
 
∴ a = 6x or 3x. 
 
If a = 3x, then MN = NL = 3x, so ∆MNL becomes isosceles and hence ∠NPL = 90°, which means that NL2 = NP2 + PL2.
 
NL2 = 9x2 and  NP2 + PL2 = 4x2 + (x2/7) = (29/7)x≠  9x2. Hence a ≠ 3x.
 
∴ a = 6x.
 
NP : NL: MN = 2x : 3x : 6x = 2 : 3 : 6.
 
Hence option (b).
         
Note: We can use the direct formula for the length of the angle bisector L. Consider the triangle drawn below

 

L = (2abCosQ)/(a + b)

So, 2x = (2 × MN × 3x × Cos60)/(MN + 3x)

Solving, we get; MN = 6x.

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