We have to prove that points (0, -1, -1), (4, 5, 1), (3, 9, 4) and (-4, 4, 4) are coplanar.

Now let us find the equation of plane passing through three point’s i.e.

(0, -1, -1), (4, 5, 1), (3, 9, 4)

By using the formula, equation of plane passing through three points is given as:

\( \begin{vmatrix} x & y+1 & z+1 \\[0.3em] 4& 6 & 2 \\[0.3em] 3 & 10 & 5 \end{vmatrix}\)

10x – 14y + 22z + 8 = 0

Divide by 2, we get

5x – 7y + 11z + 4 = 0 ……. (1)

By using the fourth point (-4, 4, 4),

Substitute the values as x = -4, y =4, z = 4 in equation (1), we get

5x – 7y + 11z + 4 = 0

5(-4) – 7(4) + 11(4) + 4 = 0

-48 + 48 = 0

0 = 0

LHS = RHS

Since, fourth point satisfies the equation of plane passing through three points. So, all the points are coplanar.

Hence, the equation of common plane is 5x – 7y + 11z + 4 = 0.

Answered by Sakshi | 1 year agoA plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (1, -2, 3). Find the equation of the plane.

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