Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).

We are given with points, (2, 1, 0), (3, -2, -2) and (3, 1, 7).

We know that, the equation of a line passing through three non-collinear points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given as

Here,

(x₁, y₁, z₁) ≡ (2, 1, 0)

(x₂, y₂, z₂) ≡ (3, -2, -2)

(x₃, y₃, z₃) ≡ (3, 1, 7)

So,

x₁ = 2, y₁ = 1 and z₁ = 0

x₂ = 3, y₂ = -2 and z₂ = -2

x₃ = 3, y₃ = 1 and z₃ = 7

Putting these values in the equation of the line, we get

Take 1^st row and 1^st column, multiply the first element of the row (a₁₁) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 1^st column.

Here,

Now take 1^st row and 2^nd column, multiply the second element of the row (a₁₂) with the difference of multiplication of opposite elements (a₂₁ × a₃₃ – a₂₃ × a₃₁), excluding 1^st row and 2^nd column.

Here,

Similarly, take 1^st row and 3^rd column, multiply the third element of the row (a₁₃) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 3^rd column.

Here,

Solving it further,

Now, since

⇒ -21x – 9y + 3z + 51 = 0

⇒ -21x – 9y + 3z = -51

⇒ -3(7x + 3y – z) = -3 × 17

⇒ 7x + 3y – z = 17

Thus, required equation of the plane is 7x + 3y – z = 17.