Question 1026678
{{{x/(b+c-a)=y/(c+a-b)=z/(a+b-c)}}} is the symmetric equation of a straight line in {{{R^3}}} with direction vector {b+c-a, c+a-b, a+b-c}.
One point on the line is (b+c-a, c+a-b, a+b-c). Another point is the origin (0,0,0).  (Why?) Incidentally, the point (b+c-a, c+a-b, a+b-c) is also contained in the plane 

x(b-c)+y(c-a)+z(a-b) = 0, 

since (b+c-a)(b-c) + (c+a - b)(c-a) + (a+b-c)(a-b)=0.  (VERIFY!!)

Hence the line {{{x/(b+c-a)=y/(c+a-b)=z/(a+b-c)}}} and the plane x(b-c)+y(c-a)+z(a-b) = 0 share two common points, namely (0,0,0) and (b+c-a,c+a-b,a+b-c), and thus the line {{{x/(b+c-a)=y/(c+a-b)=z/(a+b-c)}}} is contained completely in the plane x(b-c)+y(c-a)+z(a-b) = 0.

Therefore if (x,y,z) satisfy {{{x/(b+c-a)=y/(c+a-b)=z/(a+b-c)}}}, then (x,y,z) should also satisfy x(b-c)+y(c-a)+z(a-b) = 0, and the statement is proved.