# Find equation of the plane passing through A (1, 4, 2), B (-1, -2, 3), C (2, 1, 1).

## Question:

Find equation of the plane passing through {eq}\displaystyle A (1,\ 4,\ 2),\ B (-1,\ -2,\ 3),\ C (2,\ 1, \ 1) {/eq}.

## Equation Of A Plane

The equation of a plane is an equation which is satisfied by every point on the plane. The equation of a plane would be of the form ax+by+cz+d=0, where a,b,c and d are constants. The equation of a plane passing through the points, {eq}(x_1, y_1,z_1), (x_2, y_2,z_2) {/eq} and {eq}(x_3, y_3, z_3) {/eq} is given by the formula,

{eq}\begin{vmatrix} x-x_1 &y-y_1 &z-z_1 \\ x_2-x_1 &y_2-y_1 &z_3-z_1 \\ x_3-x_1 &y_3-y_1 &z_3-z_1 \end{vmatrix}=0 {/eq}

## Answer and Explanation:

The given points are,

{eq}(x_1,y_1,z_1)=(1,4,2); (x_2, y_2, z_2) = (-1,-2,3); (x_3,y_3,z_3)=(2,1,1) {/eq}

The equation of the required plane through these points is,

{eq}\begin{vmatrix} x-x_1 &y-y_1 &z-z_1 \\ x_2-x_1 &y_2-y_1 &z_3-z_1 \\ x_3-x_1 &y_3-y_1 &z_3-z_1 \end{vmatrix}=0 {/eq}

{eq}\left| \begin{array}{ccc}{x-1} & {y-4} & {z-2} \\ {-1-1} & {-2-4} & {3-2} \\ {2-1} & {1-4} & {1-2}\end{array}\right|=0 \\ \left| \begin{array}{ccc}{x-1} & {y-4} & {z-2} \\ {-2} & {-6} & {1} \\ {1} & {-3} & {-1}\end{array}\right|=0 \\ \begin{array}{l}{(x-1)(6+3)-(y-4)(2-1)+(z-2)(6+6)=0} \\ {9(x-1)-(y-4)+12(z-2)=0}\end{array} \\ 9 x-y+12 z-29=0 {/eq}

Therefore, the equation of the required plane is, {eq}\mathbf{9 x-y+12 z-29=0} {/eq}

#### Learn more about this topic:

Finding the Equation of a Plane from Three Points

from NDA Exam Preparation & Study Guide

Chapter 7 / Lesson 11
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