Understanding the Problem

The core task is to determine the equation of a plane in three-dimensional space. This plane must possess two specific properties:

• It must contain a given line, defined by its symmetric equations.
• It must be parallel to another given line, also defined by its symmetric equations.

Key Concepts: Lines and Planes in 3D

To solve this problem effectively, we need to understand the fundamental concepts related to lines and planes in 3D geometry:

• Line Representation: A line can be represented in symmetric form as \(\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\). This indicates the line passes through the point \(P_0(x_0, y_0, z_0)\) and has a direction vector \(\vec{d} = \langle a, b, c \rangle\).
• Plane Equation: The standard equation of a plane is \(Ax + By + Cz = D\). This plane has a normal vector \(\vec{n} = \langle A, B, C \rangle\). If the plane passes through a point \(P_0(x_0, y_0, z_0)\), its equation can also be written as \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\).
• Plane Parallel to a Line: A plane is parallel to a line if its normal vector \(\vec{n}\) is perpendicular to the direction vector \(\vec{d}\) of the line.
• Plane Containing a Line: A plane contains a line if it passes through any point on the line and its normal vector \(\vec{n}\) is perpendicular to the direction vector \(\vec{d}\) of the line.
• Finding the Normal Vector: If a plane is parallel to two non-parallel vectors, say \(\vec{u}\) and \(\vec{v}\), its normal vector \(\vec{n}\) can be found by taking the cross product: \(\vec{n} = \vec{u} \times \vec{v}\).

Step-by-Step Solution

Step 1: Extract Information from Given Lines

We are provided with the equations of two lines:

1. Line 1: \(\frac{{{\rm{x}} - 1}}{3} = \frac{{{\rm{y}} - 4}}{2} = \frac{{{\rm{z}} - 4}}{{ - 2}}\)
   • This line lies within the plane we need to find.
   • It passes through the point \(P_1(1, 4, 4)\).
   • The direction vector of this line is \(\vec{a} = \langle 3, 2, -2 \rangle\). Since the plane contains this line, \(\vec{a}\) must be parallel to the plane.
2. Line 2: \(\frac{{{\rm{x}} + 1}}{2} = \frac{{{\rm{y}} - 1}}{{ - 4}} = \frac{{{\rm{z}} + 2}}{1}\)
   • This line is parallel to the plane.
   • The direction vector of this line is \(\vec{b} = \langle 2, -4, 1 \rangle\). Since the plane is parallel to this line, \(\vec{b}\) must also be parallel to the plane.

Step 2: Calculate the Normal Vector of the Plane

The normal vector \(\vec{n}\) to the plane must be perpendicular to both \(\vec{a}\) and \(\vec{b}\), as both vectors are parallel to the plane. We can find \(\vec{n}\) by computing the cross product of \(\vec{a}\) and \(\vec{b}\):

\(\vec{n} = \vec{a} \times \vec{b}\)

Using the determinant method for the cross product:

\(\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & -2 \\ 2 & -4 & 1 \end{vmatrix}\)

Expanding the determinant:

\(\vec{n} = \mathbf{i} \left( (2)(1) - (-2)(-4) \right) - \mathbf{j} \left( (3)(1) - (-2)(2) \right) + \mathbf{k} \left( (3)(-4) - (2)(2) \right)\)

\(\vec{n} = \mathbf{i} (2 - 8) - \mathbf{j} (3 - (-4)) + \mathbf{k} (-12 - 4)\)

\(\vec{n} = \mathbf{i} (-6) - \mathbf{j} (7) + \mathbf{k} (-16)\)

\(\vec{n} = \langle -6, -7, -16 \rangle\)

We can use any scalar multiple of this vector as the normal vector. Choosing \(\vec{n'} = -\vec{n} = \langle 6, 7, 16 \rangle\) simplifies the coefficients.

Step 3: Formulate the Plane Equation

The equation of a plane passing through a point \(P_1(x_0, y_0, z_0)\) and having a normal vector \(\vec{n'} = \langle A, B, C \rangle\) is given by the formula:

\(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\)

Substituting the point \(P_1(1, 4, 4)\) and the normal vector \(\vec{n'} = \langle 6, 7, 16 \rangle\):

\(6(x - 1) + 7(y - 4) + 16(z - 4) = 0\)

Now, distribute the coefficients:

\(6x - 6 + 7y - 28 + 16z - 64 = 0\)

Combine the constant terms:

\(6x + 7y + 16z - (6 + 28 + 64) = 0\)

\(6x + 7y + 16z - 98 = 0\)

Finally, rewrite the equation in the standard form \(Ax + By + Cz = D\):

\(6x + 7y + 16z = 98\)

Step 4: Compare with Given Options

Let's check which of the provided options matches our derived equation:

Our calculated equation, \(6x + 7y + 16z = 98\), perfectly matches the equation presented in Option 3.

Conclusion

By identifying a point on the plane and calculating the normal vector using the cross product of the direction vectors of the given lines, we found the equation of the required plane. The correct equation is \(6x + 7y + 16z = 98\).