To solve the problem step by step, we will first find the point of intersection \( P \) of the line segment \( QR \) with the plane, and then we will find the foot of the perpendicular \( S \) from point \( T \) to the line \( QR \). Finally, we will calculate the length of the segment \( PS \).
### Step 1: Find the equation of the line \( QR \)
The points \( Q(2, 3, 5) \) and \( R(1, -1, 4) \) can be used to find the direction ratios of the line \( QR \):
- Direction ratios \( \vec{d} = R - Q = (1 - 2, -1 - 3, 4 - 5) = (-1, -4, -1) \).
The parametric equations of the line can be written as:
\[
\frac{x - 2}{-1} = \frac{y - 3}{-4} = \frac{z - 5}{-1} = \lambda
\]
This can be rewritten as:
\[
x = 2 - \lambda, \quad y = 3 - 4\lambda, \quad z = 5 - \lambda
\]
### Step 2: Find the intersection point \( P \) with the plane \( 5x - 4y - z = 1 \)
Substituting the parametric equations into the plane equation:
\[
5(2 - \lambda) - 4(3 - 4\lambda) - (5 - \lambda) = 1
\]
Expanding this:
\[
10 - 5\lambda - 12 + 16\lambda - 5 + \lambda = 1
\]
Combining like terms:
\[
(10 - 12 - 5) + (-5\lambda + 16\lambda + \lambda) = 1
\]
\[
-7 + 12\lambda = 1
\]
Solving for \( \lambda \):
\[
12\lambda = 8 \implies \lambda = \frac{2}{3}
\]
Now substituting \( \lambda \) back to find the coordinates of point \( P \):
\[
x = 2 - \frac{2}{3} = \frac{4}{3}, \quad y = 3 - 4\left(\frac{2}{3}\right) = 3 - \frac{8}{3} = \frac{1}{3}, \quad z = 5 - \frac{2}{3} = \frac{13}{3}
\]
Thus, the coordinates of point \( P \) are:
\[
P\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)
\]
### Step 3: Find the foot of the perpendicular \( S \) from point \( T(2, 1, 4) \) to line \( QR \)
The direction ratios of line \( QR \) are \( (-1, -4, -1) \). The vector from point \( T \) to a point on line \( QR \) can be expressed as:
\[
\vec{TS} = (x - 2, y - 1, z - 4)
\]
We want this vector to be perpendicular to the direction ratios of line \( QR \):
\[
(-1)(x - 2) + (-4)(y - 1) + (-1)(z - 4) = 0
\]
Substituting the parametric equations of the line \( QR \):
\[
(-1)(2 - \lambda - 2) + (-4)(3 - 4\lambda - 1) + (-1)(5 - \lambda - 4) = 0
\]
Simplifying:
\[
0 - 4(2 - 4\lambda) - (1 - \lambda) = 0
\]
\[
-8 + 16\lambda - 1 + \lambda = 0
\]
\[
17\lambda - 9 = 0 \implies \lambda = \frac{9}{17}
\]
Now substituting \( \lambda \) back to find the coordinates of point \( S \):
\[
x = 2 - \frac{9}{17} = \frac{34 - 9}{17} = \frac{25}{17}, \quad y = 3 - 4\left(\frac{9}{17}\right) = \frac{51 - 36}{17} = \frac{15}{17}, \quad z = 5 - \frac{9}{17} = \frac{85 - 9}{17} = \frac{76}{17}
\]
Thus, the coordinates of point \( S \) are:
\[
S\left(\frac{25}{17}, \frac{15}{17}, \frac{76}{17}\right)
\]
### Step 4: Calculate the length of the segment \( PS \)
Using the distance formula:
\[
PS = \sqrt{\left(\frac{25}{17} - \frac{4}{3}\right)^2 + \left(\frac{15}{17} - \frac{1}{3}\right)^2 + \left(\frac{76}{17} - \frac{13}{3}\right)^2}
\]
Calculating each term:
1. For \( x \):
\[
\frac{25}{17} - \frac{4}{3} = \frac{75 - 68}{51} = \frac{7}{51}
\]
2. For \( y \):
\[
\frac{15}{17} - \frac{1}{3} = \frac{45 - 17}{51} = \frac{28}{51}
\]
3. For \( z \):
\[
\frac{76}{17} - \frac{13}{3} = \frac{228 - 221}{51} = \frac{7}{51}
\]
Now substituting into the distance formula:
\[
PS = \sqrt{\left(\frac{7}{51}\right)^2 + \left(\frac{28}{51}\right)^2 + \left(\frac{7}{51}\right)^2}
\]
Calculating:
\[
PS = \sqrt{\frac{49 + 784 + 49}{2601}} = \sqrt{\frac{882}{2601}} = \frac{7\sqrt{18}}{51} = \frac{14\sqrt{2}}{51}
\]
### Final Answer
The length of the line segment \( PS \) is:
\[
\frac{14\sqrt{2}}{51}
\]
