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To find the distance from a point to a line in three-dimensional space, we use the formula:

\[
\text{Distance} = \frac{|\mathbf{AB} \cdot (\mathbf{v})|}{|\mathbf{v}|}
\]

Where:
- Point \( A = (2, 9, 6) \)
- The line is given in parametric form: \(\mathbf{a}(t) = \langle -1, 2, -4 \rangle t + \langle 4, -3, -2 \rangle\)
- Point \( B = (4, -3, -2) \) (from the constant vector in the parametric equation)
- Vector \( \mathbf{v} = \langle -1, 2, -4 \rangle \) (from the coefficient of \( t \) in the parametric equation)

Let's proceed step-by-step to calculate the distance.  

### Step 1: Find Vector AB

\[
\mathbf{AB} = \mathbf{A} - \mathbf{B} = (2 - 4, 9 - (-3), 6 - (-2)) = (-2, 12, 8)
\]

### Step 2: Compute the Cross Product \( \mathbf{AB} \times \mathbf{v} \)

\[
\mathbf{AB} \times \mathbf{v} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-2 & 12 & 8 \\
-1 & 2 & -4
\end{vmatrix}
\]

### Step 3: Find the Magnitude of the Cross Product

The magnitude of a vector \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) is given by:

\[
|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}
\]

### Step 4: Find the Magnitude of Vector \( \mathbf{v} \)

Now let's compute these steps.The distance from the point \( (2, 9, 6) \) to the line defined by \( \mathbf{a}(t) = \langle -1, 2, -4 \rangle t + \langle 4, -3, -2 \rangle \) is approximately **14.50** units.

Would you like more details on any step or have any other questions?

### Related Questions:
1. How is the cross product of two vectors calculated?
2. What is the geometric interpretation of the distance formula used here?
3. How do you find the distance between a point and a plane?
4. Can we find the shortest distance between two skew lines in 3D space?
5. How is the dot product different from the cross product?

### Tip:
When working with vectors, knowing when to use dot products (for projections and angles) versus cross products (for areas and perpendiculars) is crucial for solving geometry problems in 3D.

Find the distance from the point (2, 9, 6) to the line \(\mathbf{a}(t) = \langle -1, 2, -4 \rangle t + \langle 4, -3, -2 \rangle \).

Learn how to find the distance from the point (2, 9, 6) to the line \(\mathbf{a}(t) = \langle -1, 2, -4 \rangle t + \langle 4, -3, -2 \rangle\) using vector algebra. This problem involves calculating the cross product and magnitude of vectors and is suitable for students in Grades 11-12 or college-level geometry courses.

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