Given :

Equation of plane :

To Find :

Equation of line

Point of intersection

Formula :

Equation of line passing through point A with position vector and parallel to vector is

Where,

From the given equation of the plane

………eq(1)

The normal vector of the plane is

As the given line is perpendicular to the plane therefore will be parallel to the line.

Now, the equation of the line passing through and parallel to is

………eq(2)

This is the required equation line.

Substituting in eq(1)

⇒6x – 3y + 5z = -2 ………eq(3)

Also substituting in eq(2)

Comparing coefficients of

⇒

………eq(4)

Let Q(a, b, c) be the point of intersection of given line and plane

As point Q lies on the given line.

Therefore from eq(4)

⇒a = 6k+2, b = -3k-3, c = 5k-5

Also point Q lies on the plane.

Therefore from eq(3)

6a – 3b + 5c = -2

⇒6(6k+2) – 3(-3k-3) + 5(5k-5) = -2

⇒36k + 12 + 9k + 9 + 25k – 25 = -2

⇒70k = 2

⇒

Therefore co-ordinates of the point of intersection of line and plane are

Q ≡