1. Simple Vector sum

determine if any of the lines are equal or parallel

L1: (x-8)/4 = (y+5)/-2 = (z+9)/3

L2: (x+7)/2 = (y-4)/1 = (z+6)/5

L3: (x+4)/-8 = (y-1)/4 = (z+18)/-16

L4: (x-2)/-2 = (y+3)/1 = (z-4)/1.5

im not sure whether the sum is wrong... Because answer says L1 n L3 are equal. But?? how to prove it

2. You are using the word "sum" in a very strange way!

If you are given two lines as
$\frac{x- a}{A}= \frac{y- b}{B}= \frac{z- c}{C}$
and
$\frac{x- u}{U}= \frac{y- v}{V}= \frac{z- w}{W}$
then the vector <A, B, C> points in the direction of the first line and the vector <U, V, W> points in the direction of the second line. They are "parallel" if and only if <A, B, C> is a multiple of <U, V, W> and they are not the same line.

To compare L1 with L2, 4= 2(2) but -2 is not 2(1) so they are not parallel nor can they be the same line.

To compare L1 with L3, -8= 4(-2) and 4= 2(-2) but -16 is not 3(-2) so they are not parallel nor can they be the same line.

If that "3" were an "8" so that L1 was given by
$\frac{x- 8}{4}= \frac{y+ 4}{-2}= \frac{z+ 9}{8}$
then it and L3,
$\frac{x+4}{-8}= \frac{y-1}{4}= \frac{z+18}{-16}$
would have parallel vectors.

However, (8, -4, -9) is a point on the first line and putting x= 8, y= -4, z= 9 in the equation of the second,
$\fac{8+ 4}{-8}= -\frac{12}{8}= -\frac{3}{2}$
$\farc{-4-1}{4}= -\frac{5}{4}\ne -\frac{3}{2}$
so these lines would be parallel but NOT equal.