find the equations of the line through the point (-2, 3, 4) and is parallel to the plane 2x + 3y + 4z = 5 and 3x + 4y + 5z = 6.

To find the equations of the line that passes through the point \((-2, 3, 4)\) and is parallel to both planes \(2x + 3y + 4z = 5\) and \(3x + 4y + 5z = 6\), follow these steps:

### Step 1: Determine the direction vector of the line
The direction vector of the line must be perpendicular to the normal vectors of both planes. The normal vector of the first plane \(2x + 3y + 4z = 5\) is \(\mathbf{n}_1 = \langle 2, 3, 4 \rangle\), and the normal vector of the second plane \(3x + 4y + 5z = 6\) is \(\mathbf{n}_2 = \langle 3, 4, 5 \rangle\).

The direction vector \(\mathbf{d}\) of the line can be found by taking the cross product of \(\mathbf{n}_1\) and \(\mathbf{n}_2\):

\[
\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2
\]

### Step 2: Calculate the cross product
The cross product \(\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2\) is calculated as follows:

\[
\mathbf{d} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 4 \\
3 & 4 & 5 \\
\end{vmatrix}
\]

This expands to:

\[
\mathbf{d} = \mathbf{i} \left(3 \times 5 - 4 \times 4\right) - \mathbf{j} \left(2 \times 5 - 4 \times 3\right) + \mathbf{k} \left(2 \times 4 - 3 \times 3\right)
\]

\[
\mathbf{d} = \mathbf{i} (15 - 16) - \mathbf{j} (10 - 12) + \mathbf{k} (8 - 9)
\]

\[
\mathbf{d} = \langle -1, 2, -1 \rangle
\]

So, the direction vector of the line is \(\mathbf{d} = \langle -1, 2, -1 \rangle\).

### Step 3: Write the parametric equations of the line
The line passes through the point \((-2, 3, 4)\) and has the direction vector \(\langle -1, 2, -1 \rangle\). The parametric equations of the line can be written as:

\[
x = -2 - t
\]
\[
y = 3 + 2t
\]
\[
z = 4 - t
\]

where \(t\) is the parameter.

### Final Answer
The equations of the line are:

\[
x = -2 - t, \quad y = 3 + 2t, \quad z = 4 - t
\]

Would you like more details or have any questions?

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Here are five related questions to deepen your understanding:

1. How can you verify if a given point lies on this line?
2. What is the distance between the point \((-2, 3, 4)\) and the plane \(2x + 3y + 4z = 5\)?
3. How would the line change if it were to pass through a different point?
4. What would be the intersection point if this line intersected one of the planes?
5. How can you find the angle between this line and one of the planes?

**Tip:** Remember, the cross product of two vectors is always perpendicular to both, making it a useful tool for finding direction vectors in problems involving planes.

Learn how to find the equations of a line passing through a given point and parallel to two planes: 2x + 3y + 4z = 5 and 3x + 4y + 5z = 6. This problem involves calculating the cross product of normal vectors, determining direction vectors, and writing parametric equations of lines. Suitable for advanced high school students and beyond.

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