Apply
Case No. 8 and
Case No. 11 to find the slope at the right support.
$\theta = \dfrac{w_oL^3}{24EI} - \dfrac{ML}{3EI}$
$\theta = \dfrac{80(9^3)}{24EI} - \dfrac{3P(9)}{3EI}$
$\theta = \dfrac{2430}{EI} - \dfrac{9P}{EI}$
$3\theta = \dfrac{7290}{EI} - \dfrac{27P}{EI}$
Use Case No. 1 for the deflection at the free end due to concentrated load P.
$\delta = \dfrac{PL^3}{3EI}$
$\delta = \dfrac{P(3^3)}{3EI}$
$\delta = \dfrac{9P}{EI}$
$\delta = 3\theta$
$\dfrac{36P}{EI} = \dfrac{7290}{EI}$
$P = 202.5 \, \text{ lb}$ answer