# Write an equation for the plane that contains the points (4,0,-6), (-8,-10,4), and (0,6,-8) in the form ax+by+cz=d

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To find the equation of the plane, we need to find its normal vector. To do this, we will get the cross product of two vectors on the plane.

There are many possible vectors, but let's use:

v1 = <12, 10, -10>

v2 = <-8, -16, 12>

v1 x v2 = <-40, -64, -112> This is a normal vector of the plane.

Plugging these in to the equation in the point normal form, we have

-40(x-4) -64(y-0) -112(z+6) = 0

Simplify: -40x -64y -112z = 512 –––>**-5x -8y -14z = 64**

There are many possible vectors, but let's use:

v1 = <12, 10, -10>

v2 = <-8, -16, 12>

v1 x v2 = <-40, -64, -112> This is a normal vector of the plane.

Plugging these in to the equation in the point normal form, we have

-40(x-4) -64(y-0) -112(z+6) = 0

Simplify: -40x -64y -112z = 512 –––>