Answer:

Let us take,

The equation of the plane passing through A (3, 2, -5)

a (x – 3) + b (y – 2) + c (z + 5) = 0

It passes through the points B (-1, \4, -3) and C (-3, 8, -5)

a (1 – 3) + b (4 – 2) + c (-3 + 5) = 0

– 4a + 2b + 2c = 0

Dividing the entire equation by 2 and multiplying by negative sign,

2a – b – c = 0

a (- 3 – 3) + b (8 – 2) + c (5 – 5) = 0

– 6a + 6b + 0c = 0

Dividing the equation by 6 and multiplying by negative sign,

a – b – 0c = 0

By cross multiplying the equations,

a/-1 = b/-1 = c/-1

Multiplying by negative sign,

a/1 = b/1 = c/1

a = k, b = k, c = k

Substituting the value,

k (x – 3) + k (y – 2) + k (z + 5) = 0

Dividing the entire equation by k

(x – 3) + (y – 2) + (z + 5) = 0

x + y + z = 0

The equation of plane passing through A (3, 2, -5), B (-1, 4, -3) and C (-3, 8, -5) is x + y + z = 0

Fourth point D (-3, 2, 1) also satisfies x + y + z = 0

The given points are coplanar

Hence, the equation of the plane containing them is x + y + z = 0.