To solve the given problem step by step, we will follow the outlined approach in the video transcript.
### Step 1: Find the General Point on Line AB
The equation of line AB is given by:
\[
\frac{x}{2} = \frac{y}{-3} = \frac{z}{6} = k
\]
From this, we can express the coordinates of any point on line AB in terms of \( k \):
\[
x = 2k, \quad y = -3k, \quad z = 6k
\]
### Step 2: Determine the Coordinates of Point N
Point P is given as \( P(1, 2, 5) \). We need to find point N on line AB such that line PN is perpendicular to line AB. The coordinates of point N can be expressed as:
\[
N(2k, -3k, 6k)
\]
The direction ratios of line PN are:
\[
\text{Direction Ratios of PN} = (2k - 1, -3k - 2, 6k - 5)
\]
### Step 3: Find Direction Ratios of Line AB
The direction ratios of line AB are:
\[
(2, -3, 6)
\]
### Step 4: Use the Condition of Perpendicularity
For lines to be perpendicular, the dot product of their direction ratios must be zero:
\[
(2k - 1) \cdot 2 + (-3k - 2) \cdot (-3) + (6k - 5) \cdot 6 = 0
\]
Expanding this gives:
\[
2(2k - 1) + 9k + 6 + 36k - 30 = 0
\]
Combining like terms:
\[
49k - 26 = 0
\]
Solving for \( k \):
\[
k = \frac{26}{49}
\]
### Step 5: Calculate the Coordinates of Point N
Substituting \( k \) back into the equations for N:
\[
N\left(2 \cdot \frac{26}{49}, -3 \cdot \frac{26}{49}, 6 \cdot \frac{26}{49}\right) = \left(\frac{52}{49}, -\frac{78}{49}, \frac{156}{49}\right)
\]
### Step 6: Find the Coordinates of Point Q
Next, we need to find point Q, where line PQ is parallel to the plane \( 3x + 4y + 5z = 0 \). The normal vector to this plane is \( (3, 4, 5) \).
### Step 7: Set Up the Equation for Line PQ
The coordinates of point Q can be expressed as:
\[
Q(2\mu, -3\mu, 6\mu)
\]
The direction ratios of line PQ are:
\[
(2\mu - 1, -3\mu - 2, 6\mu - 5)
\]
### Step 8: Use the Condition of Perpendicularity for Line PQ
Using the normal vector of the plane, we set up the equation:
\[
3(2\mu - 1) + 4(-3\mu - 2) + 5(6\mu - 5) = 0
\]
Expanding this gives:
\[
6\mu - 3 - 12\mu - 8 + 30\mu - 25 = 0
\]
Combining like terms:
\[
24\mu - 36 = 0
\]
Solving for \( \mu \):
\[
\mu = \frac{3}{2}
\]
### Step 9: Calculate the Coordinates of Point Q
Substituting \( \mu \) back into the equations for Q:
\[
Q\left(2 \cdot \frac{3}{2}, -3 \cdot \frac{3}{2}, 6 \cdot \frac{3}{2}\right) = \left(3, -\frac{9}{2}, 9\right)
\]
### Step 10: Write the Equation of Line PN
The direction ratios of line PN are:
\[
\left(2k - 1, -3k - 2, 6k - 5\right)
\]
Substituting \( k = \frac{26}{49} \):
\[
\text{Direction Ratios} = \left(3, -\frac{176}{49}, -\frac{89}{49}\right)
\]
Thus, the equation of line PN is:
\[
\frac{x - 1}{3} = \frac{y - 2}{-\frac{176}{49}} = \frac{z - 5}{-\frac{89}{49}}
\]
