The problem you provided consists of two tasks related to planes and their intersection in 3D space.

Given planes:

1. $y - z = 1$
2. $x + y + 2z = 3$

These two planes intersect along a line $\epsilon$.

Task (a): Find the parametric and symmetric equations of the line $\epsilon$.

To find the parametric and symmetric equations of the line formed by the intersection of two planes, we need to:

• Solve the system of equations.
• Find a point on the line.
• Find a direction vector for the line.

Step 1: Solve the system of equations

We have:

1. $y - z = 1$
2. $x + y + 2z = 3$

From the first equation: $y = z + 1$

Substitute this into the second equation: $x + (z + 1) + 2z = 3$ $x + 3z + 1 = 3$ $x = 2 - 3z$

So, the parametric form of the line is: $x = 2 - 3t, \quad y = t + 1, \quad z = t$ where $t$ is the parameter.

Parametric equations of the line:

$x = 2 - 3t, \quad y = t + 1, \quad z = t$

Step 2: Symmetric form

To express the symmetric form, eliminate the parameter $t$ from the parametric equations: $t = z$ From $x = 2 - 3t$, we get: $t = \frac{2 - x}{3}$ From $y = t + 1$, we get: $t = y - 1$

Thus, the symmetric form is: $\frac{2 - x}{3} = z = y - 1$

Task (b): Find the equation of a third plane that contains the line $\epsilon$ and is perpendicular to the plane $x + y - 2z = 3$.

To solve this, we follow these steps:

Step 1: Find the normal vector of the given plane.

The plane $x + y - 2z = 3$ has a normal vector