Find a plane containing the point (-7,-4,-2) and the line of intersection of the planes 3x-7y+3z=19
and -x+6y+4z=5

To find the equation of a plane containing the point \((-7, -4, -2)\) and the line of intersection of the planes \(3x - 7y + 3z = 19\) and \(-x + 6y + 4z = 5\), follow these steps:

### Step 1: Find the direction vector of the line of intersection.
The line of intersection of two planes can be represented by a direction vector, which is the cross product of the normal vectors of the two planes.

1. **Normal vector of the first plane \(3x - 7y + 3z = 19\):**
   \[
   \mathbf{n}_1 = \langle 3, -7, 3 \rangle
   \]

2. **Normal vector of the second plane \(-x + 6y + 4z = 5\):**
   \[
   \mathbf{n}_2 = \langle -1, 6, 4 \rangle
   \]

3. **Cross product to find the direction vector \(\mathbf{d}\):**
   \[
   \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2
   \]
   Compute the cross product:
   \[
   \mathbf{d} = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   3 & -7 & 3 \\
   -1 & 6 & 4 \\
   \end{vmatrix}
   \]
   Expanding the determinant:
   \[
   \mathbf{d} = \mathbf{i} \left( -7(4) - 3(6) \right) - \mathbf{j} \left( 3(4) - 3(-1) \right) + \mathbf{k} \left( 3(6) - (-7)(-1) \right)
   \]
   Simplifying:
   \[
   \mathbf{d} = \mathbf{i}(-28 - 18) - \mathbf{j}(12 + 3) + \mathbf{k}(18 - 7)
   \]
   \[
   \mathbf{d} = \mathbf{i}(-46) - \mathbf{j}(15) + \mathbf{k}(11)
   \]
   \[
   \mathbf{d} = \langle -46, -15, 11 \rangle
   \]

### Step 2: Use the direction vector and the given point to write the equation of the plane.
The equation of the plane can be written in the form:
\[
a(x + 7) + b(y + 4) + c(z + 2) = 0
\]
where \(\langle a, b, c \rangle\) is a normal vector to the plane. We already know that this plane must be perpendicular to the direction vector \(\mathbf{d} = \langle -46, -15, 11 \rangle\).

Thus, the normal vector \(\langle a, b, c \rangle\) will be parallel to \(\mathbf{d}\).

We can use any scalar multiple of \(\mathbf{d}\) for the normal vector. Therefore, we can take:
\[
\mathbf{n} = \langle -46, -15, 11 \rangle
\]

### Step 3: Substitute the point and the normal vector into the plane equation.
The plane equation becomes:
\[
-46(x + 7) - 15(y + 4) + 11(z + 2) = 0
\]
Expanding this:
\[
-46x - 322 - 15y - 60 + 11z + 22 = 0
\]
Combine like terms:
\[
-46x - 15y + 11z = 360
\]

So, the equation of the plane is:
\[
46x + 15y - 11z = 360
\]

This is the required plane that contains the point \((-7, -4, -2)\) and the line of intersection of the given planes.

Would you like any more details or have any questions?

---

1. How do you find the equation of a plane given three points?
2. How do you determine if two planes are parallel?
3. What is the geometrical significance of the cross product of two vectors?
4. How can the equation of a plane be simplified further if possible?
5. How to find the angle between two planes?

**Tip:** Always verify your results by substituting known points or using vector properties to ensure correctness.

Learn how to find the equation of a plane that contains the point (-7, -4, -2) and the line of intersection of two planes: 3x - 7y + 3z = 19 and -x + 6y + 4z = 5. Detailed step-by-step solution provided, covering vectors, equations of planes, and cross product calculations.

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