Answer
$(a)\space 322\space N$
$(b)\space 209\space N$
$(c)\space 279\space N$
Work Step by Step
Please see the attached image first.
As shown in the left side figure, the force $F$ can be first resolved into two components $F_{Z},F_{P}$
We can get,
$F_{P}=Fsin54^{\circ}=(475\space N)sin54^{\circ}=384\space N$
Then $F_{P}$ can be resolved into x, y components by projection onto x,y plane.
(a) From the right side figure,
$F_{X}=F_{P}cos33^{\circ}=(384\space N)\times0.84=322\space N$
(b) From the right side figure,
$F_{Y}=F_{P}sin33^{\circ}=(384\space N)\times0.54=209\space N$
(c) From the left side figure,
$F_{Z}=Fcos54^{\circ}=(475\space N)\times0.6=279\space N$
