A

Step 1: Identify the charges and their positions.
The point charges are q1 = 20 µC located at the point (-1, 1, 1) m and q2 = 25 µC located at (3, 1, -2) m.

Step 2: Calculate the distance vector from q1 to q2.
The position vector of q2 (r2) - position vector of q1 (r1):
r = (3, 1, -2) - (-1, 1, 1) = (3 + 1, 1 - 1, -2 - 1) = (4, 0, -3) m.

Step 3: Calculate the magnitude of the distance.
The magnitude of the distance vector r is calculated as:
$$ r = \\sqrt{(4)^2 + (0)^2 + (-3)^2} = \\sqrt{16 + 0 + 9} = \\sqrt{25} = 5 ext{ m} $$

Step 4: Calculate the electrostatic force using Coulomb's Law.
Coulomb's Law states that the force between two point charges is given by:
$$ F = k \frac{|q_1 q_2|}{r^2} $$
where k = 8.99 x 10^9 N m^2/C^2, $|q_1| = 20 imes 10^{-6}$ C, $|q_2| = 25 imes 10^{-6}$ C, and $r = 5$ m.

The force magnitude is:
$$ F = 8.99 \times 10^9 \frac{(20 \times 10^{-6})(25 \times 10^{-6})}{(5)^2} $$
$$ = 8.99 \times 10^9 \frac{500 \times 10^{-12}}{25} $$
$$ = 8.99 \times 10^9 \times 20 \times 10^{-12} $$
$$ = 179.8 N $$

Step 5: Determine the unit vector.
The unit vector along the distance vector from q1 to q2 is:
$$ \hat{r} = \frac{r}{|r|} = \frac{(4, 0, -3)}{5} = \left(\frac{4}{5}, 0, -\frac{3}{5}\right) $$

Step 6: Finalize the force vector using the unit vector.
The force vector acting on q2 due to q1 is then given by the product of the force magnitude and the unit vector:
$$ F_{q2} = F \cdot \hat{r} = 179.8 N \cdot \left(\frac{4}{5}, 0, -\frac{3}{5}\right) $$
Therefore, magnitude of the electrostatic force on q2 is 179.8 N and the unit vector is \( \left(\frac{4}{5}, 0, -\frac{3}{5}\right) \). Hence, the electrostatic force on q2 is directed towards q1.