**1. Introduction:**
In this problem, we are given the Arrhenius factors (pre-exponential factors) and activation energies for two parallel reactions. We need to determine the temperature at which the two reactions will occur at the same rate.

**2. Arrhenius Equation:**
The rate constant (k) of a reaction can be calculated using the Arrhenius equation:
k = A * exp(-Ea/RT)
where:
- k is the rate constant
- A is the Arrhenius factor (pre-exponential factor)
- Ea is the activation energy
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin

**3. Mathematical Approach:**
To find the temperature at which the two reactions occur at the same rate, we can set the rate constants of the two reactions equal to each other and solve for T.

For the first reaction:
k1 = A1 * exp(-Ea1/RT)

For the second reaction:
k2 = A2 * exp(-Ea2/RT)

Setting k1 = k2, we have:
A1 * exp(-Ea1/RT) = A2 * exp(-Ea2/RT)

Taking the natural logarithm (ln) of both sides:
ln(A1) - (Ea1/RT) = ln(A2) - (Ea2/RT)

Rearranging the equation:
(Ea2 - Ea1)/R * (1/T) = ln(A1/A2)

Simplifying the equation:
1/T = (Ea2 - Ea1)/(R * ln(A1/A2))

Finally, we can solve for T by taking the reciprocal of both sides:
T = (R * ln(A1/A2))/(Ea2 - Ea1)

**4. Calculation:**
Plugging in the given values:
A1 = 10^10 sec^-1
A2 = 10^8 sec^-1
Ea1 = 150 kJ/mol
Ea2 = 75 kJ/mol
R = 8.314 J/mol·K

Calculating the temperature:
T = (8.314 J/mol·K * ln(10^10/10^8))/(75 kJ/mol - 150 kJ/mol)

Converting kJ to J and simplifying:
T = (8.314 * ln(100))/(75 - 150)
T = (8.314 * ln(100))/(-75)

Using the properties of natural logarithm:
T ≈ (8.314 * 4.605)/(75 * -1.0)
T ≈ -0.383 K

**5. Conclusion:**
The calculated temperature of -0.383 K is not physically meaningful as temperatures cannot be negative. This suggests that the two reactions will never occur at the same rate.