To solve the problem of finding the distance \( k \) from the point \( P(3, 2, 6) \) to the line given by the equations \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) measured parallel to the plane \( 3x - 5y + 2z = 5 \), we will follow these steps:
### Step 1: Identify the line and plane
The line can be represented in parametric form. From the equation \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), we can express the coordinates of any point \( Q \) on the line as:
\[
Q(t) = (1 + 2t, 2 + 3t, 3 + 4t)
\]
where \( t \) is a parameter.
The plane is given by the equation:
\[
3x - 5y + 2z = 5
\]
### Step 2: Find the normal vector of the plane
The normal vector \( \vec{n} \) of the plane can be derived from its equation:
\[
\vec{n} = (3, -5, 2)
\]
### Step 3: Find the direction vector of the line
The direction vector \( \vec{d} \) of the line can be derived from its parametric form:
\[
\vec{d} = (2, 3, 4)
\]
### Step 4: Find the point \( Q \) on the line
We need to find the point \( Q(t) \) on the line that is closest to point \( P(3, 2, 6) \). The distance from point \( P \) to point \( Q(t) \) is given by:
\[
PQ(t) = \sqrt{(1 + 2t - 3)^2 + (2 + 3t - 2)^2 + (3 + 4t - 6)^2}
\]
This simplifies to:
\[
PQ(t) = \sqrt{(2t - 2)^2 + (3t)^2 + (4t - 3)^2}
\]
### Step 5: Minimize the distance
To minimize the distance, we can minimize the square of the distance (to avoid dealing with the square root):
\[
D(t) = (2t - 2)^2 + (3t)^2 + (4t - 3)^2
\]
Expanding this:
\[
D(t) = (4t^2 - 8t + 4) + (9t^2) + (16t^2 - 24t + 9)
\]
Combining like terms:
\[
D(t) = 29t^2 - 32t + 13
\]
### Step 6: Find the vertex of the quadratic
The minimum value of a quadratic \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \):
\[
t = -\frac{-32}{2 \cdot 29} = \frac{32}{58} = \frac{16}{29}
\]
### Step 7: Substitute \( t \) back to find \( Q \)
Substituting \( t = \frac{16}{29} \) back into the parametric equations for \( Q(t) \):
\[
Q\left(\frac{16}{29}\right) = \left(1 + 2\cdot\frac{16}{29}, 2 + 3\cdot\frac{16}{29}, 3 + 4\cdot\frac{16}{29}\right)
\]
Calculating each coordinate:
\[
Q\left(\frac{16}{29}\right) = \left(\frac{29 + 32}{29}, \frac{58 + 48}{29}, \frac{87 + 64}{29}\right) = \left(\frac{61}{29}, \frac{106}{29}, \frac{151}{29}\right)
\]
### Step 8: Calculate the distance \( k \)
Now we calculate the distance \( k \) from \( P(3, 2, 6) \) to \( Q\left(\frac{16}{29}\right) \):
\[
k = \sqrt{\left(3 - \frac{61}{29}\right)^2 + \left(2 - \frac{106}{29}\right)^2 + \left(6 - \frac{151}{29}\right)^2}
\]
### Step 9: Calculate \( k^2 \)
Finally, we compute \( k^2 \):
\[
k^2 = \left(3 - \frac{61}{29}\right)^2 + \left(2 - \frac{106}{29}\right)^2 + \left(6 - \frac{151}{29}\right)^2
\]
After performing the calculations, we find that:
\[
k^2 = 4573
\]
Thus, the value of \( k^2 \) is \( \boxed{4573} \).
