Line Equation Setup

The problem asks us to find the point where a specific line intersects a given plane. First, let's define the equation of the line.

We are given a point on the line, \(P_0 = (1, -1, 2)\), and its direction ratios, \(<a, b, c> = <3, 2, 2>\).

The parametric equations of the line can be written as:

• \(x = x_0 + at \implies x = 1 + 3t\)
• \(y = y_0 + bt \implies y = -1 + 2t\)
• \(z = z_0 + ct \implies z = 2 + 2t\)

Here, \(t\) is a parameter. As \(t\) varies, the point \((x, y, z)\) traces the line.

Plane Equation

The equation of the plane is given as:

\(x + 2y + 3z = 18\)

Finding the Intersection Point

The point of intersection must satisfy both the line's equations and the plane's equation. We can find this point by substituting the parametric equations of the line into the equation of the plane.

1. Substitute the expressions for \(x\), \(y\), and \(z\) from the line's parametric equations into the plane's equation:

   \((1 + 3t) + 2(-1 + 2t) + 3(2 + 2t) = 18\)

2. Now, simplify and solve this equation for the parameter \(t\):

   \(1 + 3t - 2 + 4t + 6 + 6t = 18\)

   Combine the constant terms and the terms with \(t\):

   \((1 - 2 + 6) + (3t + 4t + 6t) = 18\)

   \(5 + 13t = 18\)

   Subtract 5 from both sides:

   \(13t = 18 - 5\)

   \(13t = 13\)

   Divide by 13:

   \(t = \frac{13}{13} \implies t = 1\)

3. We found the value of the parameter \(t\) at the point of intersection is \(t=1\). Now, substitute this value back into the parametric equations of the line to find the coordinates of the intersection point:
   • \(x = 1 + 3(1) = 1 + 3 = 4\)
   • \(y = -1 + 2(1) = -1 + 2 = 1\)
   • \(z = 2 + 2(1) = 2 + 2 = 4\)

Therefore, the point of intersection of the line and the plane is \((4, 1, 4)\).