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}