# Find the equation

Given: Point P(1, 1, 1) and the line:

To find the equation of the plane passing through the given point and given line.

To prove: plane contains line,

Formula used:

Equation of line passing through point P(x1, y1, z1):

a(x – x1) + b(y – y1) + c(z – z1) = 0

Therefore,

Equation of line passing through point P(1, 1, 1):

a(x – 1) + b(y – 1) + c(z – 1) = 0

Since a plane contains the line

So, the plane passes through the point(-3, 1, 5) and parallel to the line with direction ratios(3, -1, -5)

a(-3 – 1) + b(1 – 1) + c(5 – 1) = 0

-4a + 4c = 0

4c = 4a

a = c……………(1)

Plane parallel to the line with direction ratios(3, -1, -5)

3a – b – 5c = 0

From (1):

3c – b – 5c = 0

- b – 2c = 0

b = -2c

Now,

a(x – 1) + b(y – 1) + c(z – 1) = 0

c(x – 1) + (-2c)(y – 1) + c(z – 1) = 0

(x – 1) – 2(y – 1) + (z – 1) = 0

x – 1 – 2y + 2 + z – 1 = 0

x – 2y + z = 0

Vector equation of plane:

If this plane contains line:

Then,

(1)(1) + (-2)(-2) + (-5)(1) = 0

1 + 4 – 5 = 0

0 = 0

which is true

So, plane, x – 2y + z = 0 contain line,

Hence Proved

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