Calling
#C->a x^2+by^2+c z^2+2 f y z + 2g z x+2hx y = 0#
and #L_1 -> x/l=y/m=z/n# one of its generators, then #L_1# must satisfy
#a l^2+b m^2+c n^2+2f m n+2g n l+2h lm = 0#
now the plane #Pi-> << p-p_0, vec v >> = 0# with
#p_0 = (0,0,0)#
#p = (x,y,z)#
#vec v = (l,m,n)#
is orthogonal to #L_1#
This plane cuts #C# in two other lines #(L_2, L_3)# such that #L_2 bot L_3# if
#(a+b+c)(l^2+m^2+n^2)-C(l,m,n)=0# or
#(a+b+c)(l^2+m^2+n^2)=0# or
#a+b+c = 0# because #l^2+m^2+n^2 ne 0#
So it is the case.
Here #vec v = (l,m,n) = (1,1,-1)#
and #f =-5, g = 8,h = 3#
then solving
#{(f y z + g z x+hx y = 0),(l x+m y + n z=0):}#
we obtain #L_2, L_3# as follows
#L_2 = {(x = z/3),(y=2/3z):}#
#L_3 = {(x = 5z),(y = -4z):}#
Attached a plot showing
#L_1# red
#L_2 # blue
#L_3# green
