Physics Forces& Friction Question *10POINTS*

An explanation with steps and the answer would be appreciated, because I am lost on this question! Thank you!

You have parked your standard-transmission 3200-kg car on a steep hill and applied
the parking brake to keep the car from rolling down the hill. The hill is inclined 75°
above the horizontal. If the coefficient of static friction between the car and the road
is 0.65, what force must be exerted by the parking break to keep the car at rest?

let the force be F
force acting downward along slope = mg*sin(75)
force action normal to slope = mg*cos(75)
Force due to friction = 0.65*mg*cos(75)
Applying force balance equation along the slope,
F+ force due to friction = force acting downwards the slope
F = mg[sin(75) - 0.65*cos(75)]
m=3200, g=9.81
Then, we get, F = 25.041 kN

By F(net) = F(gravity) - F(friction) - F(brake)
=>if F(net) is zero
=>Fg = Ff + Fa
=>Fa = Fg - Ff
=>F(applied) = mgsinθ - µs x N
=>F(applied) = mgsinθ - µs x mgcosθ
=>F(applied) = 3200 x 9.8 x [sin75* - 0.65 x cos75*]
=>F(applied) = 25015.67 N