Lines in Space
Key Questions

The symmetric equation of the line with the direction vector
#vec{v}=(a,b,c)# passing through the point#(x_0,y_0,z_0)# is:#{xx_0}/a={yy_0}/b={zz_0}/c# ,where none of
#a,b# and#c# are zero.If one of
#a,b# , and#c# is zero; for example,#c=0# , then we can write:#{xx_0}/a={yy_0}/b# and#z=z_0# .If two pf
#a,b# , and#c# are zero; for example,#b=c=0# , then we can write:#y=y_0# and#z=z_0# (There is no restriction on
#x# , it can be any real number. )
I hope that this was helpful.

We want to find the parametric equations of the line L passing through the point P and parallel to a vector A.
Let us take a 3dimensional point in
#R^3# , call it#P = (x_0,y_0,z_0)# .A line L is drawn such that it passes through P and is parallel to the vector
#A = (u,v,w)# .3 parametric equations can be written which express the components:
#x = x_0 + tu#
#y = y_0 + tv# #(oo < t < oo)#
#z = z_0 + tw# As an example, with a point
#P = (2,4,6)# and vector#A = (1,3,5)# , we have the following parametric equations:#x = 2 + tu#
#y = 4 + 3v# #(oo < t < oo)#
#z = 6 + 5w# 
Answer:
For plane:
#ax+bx+cx+d = 0# and line:
#(x, y, z) = (x_0, y_0, z_0) + t(u, v, w)# these are suitable conditions:
#{ (ax_0+by_0+cz_0+d = 0), (au+bv+cw = 0) :}# Explanation:
The most general equation of a plane in three dimensional space is:
#ax+by+cz+d = 0# A line in three dimensional space can be represented in a variety of ways, but one representation that will work for any line is the parametric form:
#(x, y, z) = (x_0, y_0, z_0) + t(u, v, w)# This line will lie in the plane if and only if two distinct points of it both satisfy the equation of the plane.
Using
#t = 0# and#t = 1# , we obtain the conditions:#{ (ax_0+by_0+cz_0+d = 0), (a(x_0+u) + b(y_0 + v) + c(z_0 + w) + d = 0) :}# Subtracting the first of these from the second, we obtain the condition:
#au+bv+cw = 0# So a necessary and sufficient set of conditions is:
#{ (ax_0+by_0+cz_0+d = 0), (au+bv+cw = 0) :}# 
I'll show how to solve this problem using an example using the vector based form of a two straight lines. Bear in mind that there will be one of the following outcomes:
 a single unique point.
 no solution (if the lines do not intersect).
 infinitely many solutions (if the lines coincide).
Suppose we have;
#L_1: \ \ \ \ vec(r_1) = ((5),(2),(1)) +lamda((1),(2),(3))# #L_2: \ \ \ \ vec(r_2) = ((2),(0),(4)) +mu((1),(2),(1))# It is vital that the variable parameters (
#lamda# and#mu# ) have different symbols, as they are different parameters relative to each line.If the lines do meet then for some specific values of
#lamda# and#mu# then:#vec(r_1) = vec(r_2)# In which case we would have:
#((5),(2),(1)) +lamda((1),(2),(3)) = ((2),(0),(4)) +mu((1),(2),(1))#
# :. \ \ \ \ \ \ ((5+lamda),(22lamda),(13lamda)) = ((2+mu),(2mu),(4mu))# By comparing the coefficients of
#hat(i)# ,#hat(j)# and#hat(k)# we have three equations in two unknowns, so we can solve for#lamda# and#mu# using the first two equations and check if the third is satisfied:# hat(i) : \ \ \ \ \ \ \ \ \ \ \ 5+lamda = 2+mu " " ....... [1]#
# hat(j) : \ \ \ \ \ \ \ \ \ 22lamda = 2mu " " ....... [2] #
# hat(k) : \ \ \ \ 13lamda = 4mu \" " ....... [3] # # Eq [1] + 1/2Eq[2] # gives#=> 6=2+2mu => mu=2#
Subs#mu=2# into#Eq[1] => 5+lamda=4=>lamda=1#
Subs#mu=2# and#lamda=1# into#Eq [3]=>1+3=2=42# So we have established that if we choose
#lamda=1# and#mu=2# then we get a unique solution simultaneously satisfying all three equations.
We can then substitute
#lamda=1# into#L_1# , (equally#mu=2# into#L_2# would work) to determine the actual coordinate of intersection.#vec(r_1) = ((5),(2),(1)) ((1),(2),(3))#
# \ \ \ = ((51),(2+2),(1+3))#
# \ \ \ = ((4),(4),(2))# so in this example, the lines
#L_1# and#L_2# intersect at the coordinate#(4,4,2)# 
Since we are talking about a line in 3D, it is more appropriate to talk about its direction vector than its slope. To find the direction vector
#vec{v}# of the line passing through two points#(x_1,y_1z_1)# and#(x_2,y_2,z_2)# can be found by#vec{v}=(x_2x_1,y_2y_1,z_2z_1)# .