# Determine the cartesian equation of the plane that passes through the points (1,4,5) and (3,2,1) and is perpendicular to the plane 2x-y+z-1 = 0.

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### 1 Answer

We have to determine the equation of the plane perpendicular to the plane 2x - y + z = 1 and passing through (1, 4, 5) and (3, 2,1).

The vector between the given points is (2, -2, -4). This vector lies in the plane. Similarly the normal to the given plane (2, -1, 1) lies in the given plane. The cross product of the two is the normal to the plane that we are finding.

`(2, -2, -4)ox(2, -1, 1)` = (-6, -10, 2) = (-3, -5, 1)

Any vector (x - 1, y - 4, z - 5) is perpendicular to (-3, -5, 1)

As the cross product of perpendicular vectors is 0

=> -3(x - 1) - 5(y - 4) + (z - 5) = 0

=> -3x + 3 - 5y + 20 + z - 5 = 0

=> 3x + 5y - z = 18

**The required plane is 3x + 5y - z = 18**