Question 1001766
The person asking the question has a meaning in mind for the expressions "a cone" and "a plane parallel to a line along the side of the cone".
Not knowing him/her, I cannot read that mind, so I must make some assumptions.


A straight line along the side of the cone must go through the vertex of the cone.
Let's place that cone on a 3D x-y-z coordinate space,
or rather, I will place x-, y-, and z-axes around that cone.
I will place the vertex at the origin,
the axis of the cone along the positive y-axis, and
the "line along the side of the cone" on the x-y plane.
The cross section of the cone with the line looks kind of like this:
{{{graph(300,300,-0.5,0.5,-0.1,0.9,abs(2x)+0.01,2x-0.01)}}} .
I assume that by cone we mean an infinite lateral surface,
with no base, or a base represented by {{{y=infinity}}} , if you wish.
I am not assuming that the infinite cone extends to any {{{y<0}}} .
The equation of the line is {{{y=mx}}} .
The equation for the cone would be {{{y=m*sqrt(x^2+z^2)}}} ,
and the z-axis is the axis that we do not see,
because it comes perpendicularly out of the screen towards us,
but at every {{{y=k}}} level, the "horizontal" section of the cone is a circle,
centered at the point with {{{system(x=0,y=k,z=0)}}} , 
with radius {{{R}}}, such that {{{R^2=x^2+z^2}}}<--->{{{R=sqrt(x^2+z^2)}}} .


Now, how could we place a plane parallel to that {{{y=mx}}} line?


The line {{{y=mx}}} is part of the plane {{{system(y=mx,z=anything)}}} that is tangent to the cone's surface.
A plane parallel to that plane, {{{system(y=mx+b,z=anything)}}} ,
is a plane that I would call parallel to the line {{{y=mx}}} .
That is probably what was envisioned in the question. I assume so.
The intersection of {{{system(y=mx+b,z=anything)}}} and {{{y=m*sqrt(x^2+z^2)}}}
is obviously
{{{mx+b=m*sqrt(x^2+z^2)}}}-->{{{(mx+b)^2=m^2(x^2+z^2)}}}-->{{{m^2x^2+b^2+2mbx=m^2x^2+m^2z2)}}}-->{{{b^2+2mbx=m^2z2)}}}-->{{{2mbx=m^2z2-b^2)}}}-->{{{x=m^2z2/2mb-b^2/2mb)}}}-->{{{x=(m/2b)*z2-b/2m)}}}
and that equation does look like a parabola.
Of course, the intersection of the cone and plane is not on the x-z plane.
{{{x=(m/2b)*z2-b/2m)}}} is just the projection of that parabolic intersection on the x-z plane.
The actual intersection is slightly stretched version of that projection. 
We can keep the x-axis, but we need a new {{{X}}} axis.
That is the line {{{system(y=mx+b,z=0)}}} ,
and on that axis a {{{DELTA}}}{{{x=1}}} becomes a distance {{{DELTA}}}{{{X=m^2+1}}} ,
so the parabola is just a little stretched.


The line {{{y=mx}}} is also part of the plane {{{system(x=anything,y=anything,z=0)}}} that contains the axis of the cone.
A plane parallel to that plane, {{{system(x=anything,y=anything,z=k)}}} ,
is also a plane that I would call parallel to the line {{{y=mx}}} .
The intersection of {{{system(x=anything,y=anything,z=k)}}} and {{{y=m*sqrt(x^2+z^2)}}}
is obviously {{{y=m*sqrt(x^2+k^2)}}}-->{{{system(y>=0,y^2=m^2(x^2+k^2))}}}-->{{{system(y>=0,y^2=m^2x^2+m^2k^2)}}}-->{{{system(y>=0,y^2-m^2x^2=m^2k^2)}}}-->{{{system(y>=0,y^2/m^2k^2-x^2/k^2=1)}}} .
That is half of a hyperbola on the {{{system(x=anything,y=anything,z=k)}}} plane.


There is an infinite number of other planes that I could consider,
but I think of them as linear combinations of the planes already described.