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.