of the vector 
is
(d) 3.
.
and 
is (b) 
.
, where 
is the
angle between 
and 
. Now
So 
. Thus 
.
and 
is
(e) 
.
So 
. Therefore
.
of
and 
is
.
through
the point 
is (a) 
.
must be proportional to
and the relation 
must be satisfied.
through the point
P=(2,4,6) and containing the line 
in parametric equation
x=2-3t, y=3+4t, z=5+2t is
and 
are parallel to the plane and hence their cross product
is normal to the plane . Thus can be described by an
equation of the form a(x-2)+b(y-4)+c(z-6)=0, where is
proportional to 
.
which is the intersection of the planes
and
can be written in parametric equation as
.
is of the form
, where the vector 
is in the direction of
and 
is a point on . Clearly
is perpendicular to normal vectors 
and
of 
and 
respectively.
So is proportional to
Thus, must be proportional to 
The more direct way of doing this is as follows:
Solve the two equations , , simultaneously and eliminate the variable, x, say. You'll get the one equation in y, z given by
You still need an equation which relates the x to either one of the two variables, y, z. So, let's solve for z in the second equation, . Then we get z = 1-2x-2y. But 5z = 3 - 8y. Combining the last two then gives
These two equations (1) and (2) define the required line. We can parametrize it by setting x=t, say, then y = 1-5x = 1-5t, by (2), and 
. That's all!