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To solve this problem, we need to find the area of the parallelogram formed by the vectors \( \vec{PQ} \) and \( \vec{PR} \).

### Step 1: Find the vectors \( \vec{PQ} \) and \( \vec{PR} \)

Given:
- \( P = (-1, 2, -6) \)
- \( Q = (0, -1, 0) \)
- \( R = (-4, 4, -6) \)

Now, compute the vectors:

1. **\( \vec{PQ} = Q - P \)**:
\[
\vec{PQ} = (0 - (-1), -1 - 2, 0 - (-6)) = (1, -3, 6)
\]

2. **\( \vec{PR} = R - P \)**:
\[
\vec{PR} = (-4 - (-1), 4 - 2, -6 - (-6)) = (-3, 2, 0)
\]

### Step 2: Find the cross product \( \vec{PQ} \times \vec{PR} \)

To find the area of the parallelogram, we need to compute the magnitude of the cross product of these vectors. The cross product of two vectors \( \vec{A} = (a_1, a_2, a_3) \) and \( \vec{B} = (b_1, b_2, b_3) \) is:

\[
\vec{A} \times \vec{B} = \left( \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}, -\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}, \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \right)
\]

Substitute the components of \( \vec{PQ} = (1, -3, 6) \) and \( \vec{PR} = (-3, 2, 0) \):

\[
\vec{PQ} \times \vec{PR} = \left( \begin{vmatrix} -3 & 6 \\ 2 & 0 \end{vmatrix}, -\begin{vmatrix} 1 & 6 \\ -3 & 0 \end{vmatrix}, \begin{vmatrix} 1 & -3 \\ -3 & 2 \end{vmatrix} \right)
\]

Now calculate the determinants:
- First component: \( (-3)(0) - (6)(2) = -12 \)
- Second component: \( -( (1)(0) - (6)(-3)) = -(0 + 18) = -18 \)
- Third component: \( (1)(2) - (-3)(-3) = 2 - 9 = -7 \)

Thus:
\[
\vec{PQ} \times \vec{PR} = (-12, -18, -7)
\]

### Step 3: Find the magnitude of the cross product

The magnitude of the cross product gives the area of the parallelogram. The magnitude of a vector \( \vec{V} = (v_1, v_2, v_3) \) is:

\[
|\vec{V}| = \sqrt{v_1^2 + v_2^2 + v_3^2}
\]

Substitute the components of \( \vec{PQ} \times \vec{PR} = (-12, -18, -7) \):

\[
|\vec{PQ} \times \vec{PR}| = \sqrt{(-12)^2 + (-18)^2 + (-7)^2} = \sqrt{144 + 324 + 49} = \sqrt{517}
\]

Thus, the area of the parallelogram is \( \sqrt{517} \), which is approximately **22.73**.

Would you like further details or have any other questions?

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### Relative Questions:
1. How do you calculate the area of a triangle formed by three points in space?
2. What is the physical significance of the cross product in vector algebra?
3. Can the area of a parallelogram be zero, and if so, under what conditions?
4. How does the cross product differ from the dot product in terms of applications?
5. Can you explain the geometric interpretation of the magnitude of the cross product?

### Tip:
For visualizing 3D geometry problems, sketching the vectors and their cross product can greatly enhance understanding!

Find the area of a parallelogram formed by the vectors PQ and PR, where P = (-1, 2, -6), Q = (0, -1, 0) and R = (-4, 4, -6).

This page explains how to calculate the area of a parallelogram formed by the vectors PQ and PR, with step-by-step details on vector subtraction, cross product, and calculating the magnitude of the result. Given points are P = (-1, 2, -6), Q = (0, -1, 0), and R = (-4, 4, -6). Suitable for high school and college-level vector algebra students.

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