# 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

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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).**