The correct increasing order for bond angles among $$BF_3$$, $$PF_3$$ and $$ClF_3$$ is :

The bond angle in a molecule depends mainly on two factors, (i) the total number of electron pairs around the central atom (bond pairs + lone pairs) and (ii) the number of lone pairs, because lone-pair-lone-pair repulsion $$\gt$$ lone-pair-bond-pair $$\gt$$ bond-pair-bond-pair repulsion (VSEPR theory).

Case 1: $$BF_3$$
• Central atom B has 3 valence electrons.
• It forms three $$\sigma$$ bonds with F and has no lone pair  $$\Rightarrow$$  steric number = 3 (AX$$_3$$ type).
• Geometry: trigonal planar.
• Ideal bond angle for trigonal planar = $$120^{\circ}$$, and with no lone pair the actual angle remains $$\approx 120^{\circ}$$.

Case 2: $$PF_3$$
• Central atom P has 5 valence electrons.
• It forms three $$\sigma$$ bonds with F and retains one lone pair  $$\Rightarrow$$  steric number = 4 (AX$$_3$$E type).
• Electron-pair geometry: tetrahedral; molecular shape: trigonal pyramidal.
• Presence of one lone pair compresses the bond angle below the ideal $$109.5^{\circ}$$. Furthermore, highly electronegative F pulls electron density away from the P-F bond pairs, so bond-pair-bond-pair repulsion decreases and the angle contracts further.
• Observed P-F-P angle $$\approx 97^{\circ}$$.

Case 3: $$ClF_3$$
• Central atom Cl has 7 valence electrons.
• It forms three $$\sigma$$ bonds with F and retains two lone pairs  $$\Rightarrow$$  steric number = 5 (AX$$_3$$E$$_2$$ type).
• Electron-pair geometry: trigonal bipyramidal; molecular shape: T-shaped (the two lone pairs occupy two equatorial positions).
• Two lone pairs produce strong repulsions, squeezing the F-Cl-F bond angle well below the ideal $$90^{\circ}$$ of a T shape; experimental value $$\approx 87^{\circ}$$.

Comparing the three angles:
$$\angle F\!-\!Cl\!-\!F \ (\text{in } ClF_3) \approx 87^{\circ} \lt$$
$$\angle F\!-\!P\!-\!F \ (\text{in } PF_3) \approx 97^{\circ} \lt$$
$$\angle F\!-\!B\!-\!F \ (\text{in } BF_3) \approx 120^{\circ}$$

Therefore, the correct increasing order of bond angles is
$$ClF_3 \lt PF_3 \lt BF_3$$.

The order matches Option B.