Answer :

**Given:** the coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, -4, 3)

**To find:** the equation of the plane

As it is given that the foot of the perpendicular drawn from origin O to the plane is P(12, -4, 3)

This means that the required plane is passing through P(12, -4, 3) and is perpendicular to OP. Let the position vector of this point P be

And it is also given the plane is normal to the line joining the points O(0,0,0) and P(12, -4, 3).

Then

Position vector of - position vector of

We know that the vector equation of a plane passing through the point and perpendicular/normal to the vector is given by

Substituting the values from eqn(i) and eqn(ii) in the above equation, we get

(by multiplying the two vectors using the formula )

is the vector equation of a required plane.

Let

Then, the above vector equation of the plane becomes,

Now multiplying the two vectors using the formula, we get

This is the Cartesian form of the equation of the required plane.

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