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To find the distance from the point \( P(6, 2, 4) \) to the line given by the parametric equations:
\[
x = 3 - t, \quad y = 6 + 4t, \quad z = 2 + 3t,
\]
we can follow this procedure:

### 1. Vector Form of the Line:
The parametric equations describe a line. To write the line in vector form, we have:
\[
\mathbf{r}(t) = \langle 3, 6, 2 \rangle + t \langle -1, 4, 3 \rangle.
\]
Thus, the line has direction vector \( \mathbf{d} = \langle -1, 4, 3 \rangle \) and passes through the point \( Q(3, 6, 2) \).

### 2. Vector from Point on Line to Given Point:
Let \( \mathbf{PQ} \) be the vector from the point \( P(6, 2, 4) \) to the point \( Q(3, 6, 2) \) on the line. This vector is:
\[
\mathbf{PQ} = P - Q = \langle 6 - 3, 2 - 6, 4 - 2 \rangle = \langle 3, -4, 2 \rangle.
\]

### 3. Project \( \mathbf{PQ} \) onto the Line Direction:
To find the perpendicular distance, we need to project \( \mathbf{PQ} \) onto the direction vector \( \mathbf{d} = \langle -1, 4, 3 \rangle \). The formula for the projection of a vector \( \mathbf{v} \) onto another vector \( \mathbf{d} \) is:
\[
\text{proj}_{\mathbf{d}} \mathbf{PQ} = \frac{\mathbf{PQ} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}.
\]
First, compute the dot products:
\[
\mathbf{PQ} \cdot \mathbf{d} = (3)(-1) + (-4)(4) + (2)(3) = -3 - 16 + 6 = -13,
\]
\[
\mathbf{d} \cdot \mathbf{d} = (-1)^2 + 4^2 + 3^2 = 1 + 16 + 9 = 26.
\]
Thus, the projection is:
\[
\text{proj}_{\mathbf{d}} \mathbf{PQ} = \frac{-13}{26} \mathbf{d} = \frac{-1}{2} \langle -1, 4, 3 \rangle = \langle \frac{1}{2}, -2, -\frac{3}{2} \rangle.
\]

### 4. Find the Perpendicular Vector:
The vector from \( P \) to the line is the difference between \( \mathbf{PQ} \) and the projection:
\[
\mathbf{v}_{\perp} = \mathbf{PQ} - \text{proj}_{\mathbf{d}} \mathbf{PQ} = \langle 3, -4, 2 \rangle - \langle \frac{1}{2}, -2, -\frac{3}{2} \rangle = \langle \frac{5}{2}, -2, \frac{7}{2} \rangle.
\]

### 5. Distance Calculation:
The distance from the point to the line is the magnitude of the perpendicular vector \( \mathbf{v}_{\perp} \):
\[
d = \left| \mathbf{v}_{\perp} \right| = \sqrt{\left( \frac{5}{2} \right)^2 + (-2)^2 + \left( \frac{7}{2} \right)^2} = \sqrt{\frac{25}{4} + 4 + \frac{49}{4}} = \sqrt{\frac{74}{4} + 4} = \sqrt{\frac{90}{4}} = \frac{\sqrt{90}}{2}.
\]
Simplifying:
\[
d = \frac{3\sqrt{10}}{2}.
\]

So, the distance from the point to the line is \( \frac{3\sqrt{10}}{2} \).

Would you like more details or have any questions?

Here are some related questions you might explore:
1. How do you find the parametric equations of a line given two points?
2. How is the formula for projecting one vector onto another derived?
3. What are the different methods to find the distance between a point and a plane?
4. How do you calculate the shortest distance between two skew lines?
5. How do you determine if a point lies on a given line in 3D space?

**Tip:** Always verify if the projection vector lies in the same direction as the line, as this ensures you're calculating the correct perpendicular distance.

Find the distance from the point to the line: (6, 2, 4); x = 3 - t, y = 6 + 4t, z = 2 + 3t

Explore how to calculate the distance between the point (6, 2, 4) and the parametric line given by the equations x = 3 - t, y = 6 + 4t, z = 2 + 3t. This problem applies concepts of vector projection and distance calculation in 3D space, suitable for advanced high school or early college students.

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