Consider the equation of line AB is `(x)/(2)=(y)/(-3)=(z)/(6)`. Through a point P(1, 2, 5) line PN is drawn perendicular to AB and line PQ is drawn parallel to the plane `3x+4y+5z=0` to meet AB is Q. Then,

Options

A. coordinate of N are `((52)/(49), -(78)/(49), (156)/(49))`
B. the coordinate of Q are `(3, -(9)/(2), 9)`
C. the equation of PN is `(x-1)/(3)=(y-2)/(-176)=(z-5)/(-89)`
D. coordinate of N are `((156)/(49), (52)/(49), -(78)/(49))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Equation of Line AB
The equation of line AB is given as:
\[ \frac{x}{2} = \frac{y}{-3} = \frac{z}{6} \]
This can be expressed in parametric form as:
\[ x = 2t, \quad y = -3t, \quad z = 6t \]
where \( t \) is a parameter.

### Step 2: Find the Coordinates of Point N
Let point N on line AB be represented by the parameter \( \lambda \):
\[ N(2\lambda, -3\lambda, 6\lambda) \]
We need to find the coordinates of N such that line PN is perpendicular to line AB.

### Step 3: Determine Direction Ratios of Line PN
The coordinates of point P are given as \( P(1, 2, 5) \). The direction ratios of line PN can be calculated as:
\[ \text{Direction ratios of PN} = (1 - 2\lambda, 2 + 3\lambda, 5 - 6\lambda) \]

### Step 4: Direction Ratios of Line AB
The direction ratios of line AB are:
\[ (2, -3, 6) \]

### Step 5: Set Up the Perpendicularity Condition
For lines PN and AB to be perpendicular, the dot product of their direction ratios must equal zero:
\[ (1 - 2\lambda) \cdot 2 + (2 + 3\lambda) \cdot (-3) + (5 - 6\lambda) \cdot 6 = 0 \]
Expanding this:
\[ 2(1 - 2\lambda) - 3(2 + 3\lambda) + 6(5 - 6\lambda) = 0 \]
This simplifies to:
\[ 2 - 4\lambda - 6 - 9\lambda + 30 - 36\lambda = 0 \]
Combining like terms:
\[ 26 - 49\lambda = 0 \]
Thus, we find:
\[ \lambda = \frac{26}{49} \]

### Step 6: Calculate Coordinates of Point N
Substituting \( \lambda = \frac{26}{49} \) into the coordinates of 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 7: Find the Coordinates of Point Q
Let point Q on line AB be represented by the parameter \( \mu \):
\[ Q(2\mu, -3\mu, 6\mu) \]
The direction ratios of line PQ are:
\[ (1 - 2\mu, 2 + 3\mu, 5 - 6\mu) \]

### Step 8: Set Up the Parallel Condition with the Plane
The equation of the plane is given by:
\[ 3x + 4y + 5z = 0 \]
The normal vector to this plane is \( (3, 4, 5) \). For line PQ to be parallel to the plane, the dot product of the direction ratios of PQ and the normal vector must equal zero:
\[ 3(1 - 2\mu) + 4(2 + 3\mu) + 5(5 - 6\mu) = 0 \]
Expanding this:
\[ 3 - 6\mu + 8 + 12\mu + 25 - 30\mu = 0 \]
Combining like terms:
\[ 36 - 24\mu = 0 \]
Thus, we find:
\[ \mu = \frac{36}{24} = \frac{3}{2} \]

### Step 9: Calculate Coordinates of Point Q
Substituting \( \mu = \frac{3}{2} \) into the coordinates of Q:
\[ Q(2 \cdot \frac{3}{2}, -3 \cdot \frac{3}{2}, 6 \cdot \frac{3}{2}) = (3, -\frac{9}{2}, 9) \]

### Step 10: Write the Equation of Line PN
Using the coordinates of points P and N, the equation of line PN can be expressed as:
\[ \frac{x - 1}{\frac{52}{49} - 1} = \frac{y - 2}{-\frac{78}{49} - 2} = \frac{z - 5}{\frac{156}{49} - 5} \]

### Summary of Results
- Coordinates of point N: \( \left(\frac{52}{49}, -\frac{78}{49}, \frac{156}{49}\right) \)
- Coordinates of point Q: \( (3, -\frac{9}{2}, 9) \)