Find the intersection of the line through P(3,1,-2) and parallel to x-3/5=2-y/1=z+2/3, with plane x+3y-z-7 = 0

The line that is passing through the point A(3,1,-2) and
it is parallel to (x-3)/5 = (y-2)/-1 = (z+2)/3 is the
following:

(x-xA)/-5 = (y - yA)/1 = (z - zA)/-3 (we've
changed the signes of the denominators of the line (x-3)/5 = (y-2)/-1 =
(z+2)/3).

We'll replace xA,yA,zA by the given coordinates
and we'll get the equation of the parallel line:

(x-3)/-5 =
(y - 1)/1 = (z + 2)/-3

Now, we'll determine the
intersection point between the parallel line and the plane x+3y-z-7 = 0. For this
reason, we'll have to solve the system formed from the equation of the line and equation
of the plane.

First, we need to write the parametric form
of the equation of the line (x-3)/-5 = (y - 1)/1 = (z +
2)/-3.

x = 3 - 5t (1)

y = 1 +
t (2)

z = -2 - 3t (3)

Now,
we'll substitute x,y,z from the equation of the plane by the expressions (1), (2),
(3).

x+3y-z-7 = 0 <=> (3-5t) + 3(1+t) -
(-2-3t) - 7 = 0

We'll remove the
brackets:

3 - 5t + 3 + 3t + 2 + 3t - 7 =
0

We'll combine like terms:

t
+ 1 = 0

t = -1

Now, we'll
determine the coordinates of the intercepting point:

x = 3
- 5t=>x = 3 + 5 => x = 8

y = 1 + t =>
y = 1 - 1 => y = 0

z = -2-3t => z = -2+3
=> z = 1

The intercepting point of the
line passing through the point (3,1,-2) and parallel to the line (x-3)/5 = (y-2)/-1 =
(z+2)/3, with the plane x+3y-z-7 = 0 is: M(8 , 0 ,
1).