Determine whether or not( (x+y+z)==(p+q+r) and E1==E2 and (x!=p or y!=q or z!=r))

Question

there are two number E1 and E2

$$E1=Ax + By + Cz \quad\mbox{and}\quad E2=Ap + Bq + Cr$$

Value of $A,B,C$ are different and positive integers.

Value of $x,y,z,p,q,r$ may be same and they are positive integers.

can we have any value for $A,B,C,x,y,z,p,q,r$ such that following condition will be true ?

$((x+y+z)==(p+q+r ) and E1==E2$ and $(x\neq p$ or $y \neq q$ or $z\neq r) )$

Answer 1

what you get is next two equations $$Ax+By+Cz=Ap+Bq+Cr\\ x+y+z=p+q+r$$ Imagine $A,B,C$ are fixed. You get two equations for 6 variables. You can choose 4 them as you want, and the other two will be determined by the two equations.

So, answering your question: Yes, it is always possible