1. Set up the geometry

Since $AB = BC = 13\,\text{m}$ and $A,B,C$ are collinear, place them on the $x$-axis as:

$A(-13,0),\quad B(0,0),\quad C(13,0)$

The particle $P$ is on the perpendicular bisector of $AC$ at a distance $13\,\text{m}$ from $B$, so:

$P(0,13)$

All four particles have mass:

$m = 100\,\text{kg}$

2. Find distances from $P$ to $A,B,C$

• Distance $PB$:

$PB = 13\,\text{m}$

• Distance $PA = PC$:

Using Pythagoras,

$PA = PC = \sqrt{13^2 + 13^2} = 13\sqrt{2}\,\text{m}$

3. Force on $P$ due to particle $B$

Magnitude:

$F_B = \frac{Gm^2}{(13)^2}$

Since $m=100$ kg,

$F_B = \frac{G(100)^2}{169} = \frac{10000G}{169} \approx 59.17G$

This force acts vertically downward.

4. Force on $P$ due to particles $A$ and $C$

Each has magnitude:

$F_A = F_C = \frac{Gm^2}{(13\sqrt{2})^2} = \frac{G(100)^2}{338} = \frac{10000G}{338} \approx 29.59G$

These two forces are symmetric. Their horizontal components cancel.

Each makes a $45^\circ$ angle with the vertical, so vertical component of each is:

$F_{A,y} = F_A\cos 45^\circ = \frac{29.59G}{\sqrt{2}} \approx 20.93G$

Similarly,

$F_{C,y} \approx 20.93G$

Hence total vertical force due to $A$ and $C$ is:

$F_{AC} = 20.93G + 20.93G = 41.86G$

This is also downward.

5. Net gravitational force on $P$

All surviving components are downward, so:

$F = F_B + F_{AC}$

$F \approx 59.17G + 41.86G = 101.03G$

So approximately,

$F \approx 100G$

6. Check options

• A: $21G$ ❌
• B: $100G$ ✅
• C: $59G$ ❌
• D: $42G$ ❌

Therefore, the correct option is B.