Finding the Equation of a Plane Through Two Points and Parallel to the X-Axis

To find the equation of a plane that passes through the points ((1, 2, -3)) and ((2, -2, 1)) and is parallel to the X-axis, we need to follow a series of steps. Let's break this down:

Step 1: Determine the Direction Vector

A plane parallel to the X-axis has a normal vector with no X-component. First, we need to find the direction vector between the two points.

The direction vector (mathbf{d} langle 2 - 1, -2 - 2, 1 - (-3) rangle) simplifies to (mathbf{d} langle 1, -4, 4 rangle).

Step 2: Determine the Normal Vector

Since the plane is parallel to the X-axis, the normal vector (mathbf{n}) must be orthogonal to the X-axis. The X-axis is defined by the vector (mathbf{i} langle 1, 0, 0 rangle). Therefore, the normal vector (mathbf{n}) is of the form (mathbf{n} langle 0, a, b rangle).

To find suitable values for (a) and (b), we use the fact that (mathbf{n}) is orthogonal to (mathbf{d}). We solve the dot product: (mathbf{d} cdot mathbf{n} 0).

(langle 1, -4, 4 rangle cdot langle 0, a, b rangle 0)

This results in: (-4a 4b 0)

Solving for (a) and (b), we get: (b a)

So, we can choose (a 1) and (b 1), giving us the normal vector (mathbf{n} langle 0, 1, 1 rangle).

Step 3: Plane Equation

The general equation of a plane is given by (-x_0(n_x) y - y_0(n_y) z - z_0(n_z) 0).

Using the point ((1, 2, -3)), the equation of the plane is:

(-1(0) y - 2(1) z 3(1) 0)

This simplifies to:

(y - 2z 3 0)

So the equation of the plane is:

(y - z 1 0)

Conclusion

Therefore, the equation of the plane through the points ((1, 2, -3)) and ((2, -2, 1)) that is parallel to the X-axis is:

(y - z 1 0)

Some additional insights:

Identifying the Existence of Such a Plane

Let's name the two given points as (P(1, 2, -3)) and (Q(2, -2, 1)).

Infinitely many planes can be drawn through (P) and (Q), but not all of them will be parallel to the X-axis. For a plane to be parallel to the X-axis, it must be equidistant from the X-axis throughout. This means all points on the plane must be equidistant from the X-axis. However, (P) and (Q) are not equidistant from the X-axis, being 1 unit and 2 units away, respectively.

Therefore, no plane can exist parallel to the X-axis that contains both (P) and (Q).

However, some planes housing (P) and (Q) may exist which are skewed with the X-axis, meaning the planes do not intersect with the X-axis.

To summarize, the equation of the plane through (1, 2, -3) and (2, -2, 1) that is parallel to the X-axis is:

(boxed{y - z 1 0})