#Question (i)#
If #ABC# is a straight line, this means that #A, B " and " C# are colinear then
#vec(AB)=kvec(BC)# where #k in RR#
#((2-0),(5-2),(-2-(-3)))=k*((3-2),(p-5),(q+2))#
#((2),(3),(1))=k*((1),(p-5),(q+2))#
#{(k=2),(3=2p-10),(1=2q+4):}#
#<=>#, #{(k=2),(p=13/2),(q=-3/2):}#
#Question (ii)#
If the angle #hat(BAC)=90^@#, then
The dot product is
#vec(AB).vec(AC)=0#
#((2-0),(5-2),(-2-(-3))).((3-0),(p-2),(q-(-3)))=0#
#((2),(3),(1)).((3),(p-2),(q+3))=0#
#2*3+3(p-2)+(q+3)=0#
#6+3p-6+q+3=0#
#3p+q=-3#
*#Question (iii)#*
#vec(AB)=((2),(3),(1))#
#vec(AC)=((3),(p-2),(q+3))#
If #AB=AC#, then
#sqrt(2^2+3^2+1^2)=sqrt(3^2+(p-2)^2+(q+3)^2)#
#14=9+(p-2)^2+(q+3)^2#
#(p-2)^2+(q+3)^2=5#
If #p=3#, then
#1+(q+3)^2=5#
#q^2+6q+10-5=0#
#q^2+6q+5=0#
Solving this quadratic equatio in #q#
#q=(-6+-sqrt(36-20))/(2)=(-6+-sqrt16)/2#
#=-3+-2#
The possible values of #q# are
#S={-1, -5}#