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# Find an equation of the plane. The plane through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1)?

**Solution:**

Given three points (2,1, 2), (3, -8, 6) and (-2, -3, 1)

Vector equation of a plane passing through three points is

(\(\overrightarrow r - \overrightarrow a).{(\overrightarrow b - \overrightarrow a) × (\overrightarrow c - \overrightarrow a)} =0\)

A(2, 1, 2) = 2i + j + 2k = a

B(3, -8, 6) = 3i - 8j + 6k = b

C(-2, -3, 1) = -2i - 3j + k = c

(b - a) = (3i - 8j + 6k) - (2i + j + 2k)

= i - 9j + 4k

(c - a) = (-2i - 3j + k) - (2i + j + 2k)

= -4i - 4j- k

Finding the cross product of the vectors, we get

{(b - a) × (c - a)} = 25i -15j + 40k

(r - a) = (xi + yj+ zk) - (2i + j + 2k)

= (x - 2)i + (y - 1)j + (z - 2)k

Finding the dot product of the vectors, we get

(r - a). {(b - a) × (c - a)} = {(x - 2)i + (y - 1)j + (z - 2)k}.{25i -15j + 40k}

= 25(x - 2) - 15(y - 1) + 40(z - 2)

= 25x - 50 - 15y + 15 + 40z - 80

= 25x - 15y + 40z - 115

Required equation 25x - 15y + 40z - 115 = 0

## Find an equation of the plane. The plane through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1)?

**Summary:**

The equation of the plane. The plane through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1) is 25x - 15y + 40z - 115 = 0.

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