# Determine if the line of intersection of the planes x-3y+7=0 and x-2z-4=0 is parallel, perpendicular or neither to the plane 3x-6y-2z-12=0.

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We'll determine first the intercepting line of the planes x-3y+7=0 and x-2z-4=0.

For this reason, we'll find out x from both equations of the planes:

x-3y+7=0 => x = 3y - 7 => x = (y - 7/3)/(1/3) (1)

x-2z-4=0 => x = 2z + 4 => x = (z + 2)/(1/2) (2)

We'll equate (1) and (2):

x = (y - 7/3)/(1/3) = (z + 2)/(1/2)

The equation of the intercepting line is: x = (y - 7/3)/(1/3) = (z + 2)/(1/2)

Now, we'll write the parametric form of this equation:

x = t

y = 7/3 + t/3

z = -2 + t/2

We'll verify if the intercepting line is perpendicular to the plane 3x-6y-2z-12=0.

For this reason, we'll have to prove that the director vector of the intercepting line is collinear to the normal vector n, to the plane 3x-6y-2z-12=0.

The director vector of the line is v( 1 , 1/3 , 1/2).

The normal vector to the plane 3x-6y-2z-12=0 is n(3 , -6 , -2).

The vectors v and n are said to be collinear, if v = a*n

( 1 , 1/3 , 1/2) = (3a , -6a , -2a)

3a = 1 => a = 1/3

1/3 = -6a => a = -1/18

It is obvious that the values of a are different, therefore the vectors v and n are not collinear, so the intercepting line is not perpendicular to the plane 3x-6y-2z-12=0.

If the intercepting line is parallel to the plane 3x-6y-2z-12=0, then the dot product of the vectors n and v is cancelling.

v*n = 1*3 + (1/3)*(-6) + (1/2)*(-2)

v*n = 3 - 2 - 1

v*n = 0

**We notice that the dot product of the vectors n and v is cancelling, therefore, the intercepting line of the planes x-3y+7=0 and x-2z-4=0 is parallel to the plane 3x-6y-2z-12=0.**