**Equation of the Plane through a given Line**

a) If equation of the is given in symmetrical form as (x – x₁)/l = (y – y₁)/m = (z – z₁)/n. Then equation of plane is a (x – x₁) + b (y – y₁) + c (z – z₁) = 0 where a, b, c is given by al + bm + cn = 0.

b) If equation of line is given in general form as a₁x + b₁y + c₁z + d₁ = 0 = a₂x + b₂y + c₂z + d₂, then the equation of plane passing through this line is (a₁x + b₁y + c₁z + d₁) + λ (a₂x + b₂y + c₂z + d₂) = 0.

c) If the plane pass trough parallel lines r = a +λb and r = c + b, then equation of the required plane is [r – a c – a b] = 0.

**Example: **The equation of the plane passing through the point (0, 7, -7) and containing the line (x + 1)/-3 = (y – 3)/2 = (z+2)/1

**Solution: **Given that (x + 1)/-3 = (y – 3)/2 = (z + 2)/1

And point (0, 7, -7)

Any plane passing through (0, 7, -7)

a (x – 0) + b (y – 7) + c (z + 7) = 0 … (1)

If equation (1) contains the given line, it must pass through the point (-1, 3, -2) and must be parallel to this line.as equation (1) passes through (-1, 3, -2)

a (-1 – 0) + b (3 – 7) + c (-2 + 7) = 0

a + 4b – 5c = 0 … (ii)

Equation (1) parallel to the given line

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

-3a + 2b + c = 0 … (iii)

From equation (1) and (ii)

\(\frac{a}{14}=\frac{b}{14}=\frac{c}{14}\)

a = b = c = k

putting a = k, b = k, c = k in equation (i) we get

kx + k (y – 7) + k (z + 7) = 0

x + y + z = 0.