From FBD of beam CD

$\Sigma F_V = 0$

$R_C + R_3 = P$

$R_C + 0.5R_C = 960$

$R_C = 640 \, \text{ lb}$

$R_3 = 0.5(640) = 320 \, \text{ lb}$ *answer*

$\Sigma M_C = 0$

$12R_3 = 960x$

$12(320) = 960x$

$x = 4 \, \text{ ft}$

Thus, P is 8 ft to the left of D. *answer*

From the figure above, R_{c} is at the midspan of AB to produce equal reactions R_{1} and R_{2}. Thus, R_{2} and R_{3} are 6 ft apart. *answer*

From FBD of beam AB

$R_1 = 0.5(640) = 320 \, \text{ lb}$ *answer*

$R_2 = 0.5(640) = 320 \, \text{ lb}$ *answer*