Let the point P (5,9,3) lie on the top of Qutub Minar , Delhi. Find the image of the point on the line :
`(x-1)/(2) = (y-2)/(3) = (z -3)/(4)` .

Answer

To find the image of the point P(5, 9, 3) on the given line, we will follow these steps:

### Step 1: Write the parametric equations of the line
The line is given in the symmetric form:
\[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda \]
From this, we can derive the parametric equations:
\[ x = 2\lambda + 1 \]
\[ y = 3\lambda + 2 \]
\[ z = 4\lambda + 3 \]

### Step 2: Define the point B on the line
Let B be a point on the line corresponding to the parameter \(\lambda\). The coordinates of point B can be expressed as:
\[ B(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \]

### Step 3: Find the direction vector of the line
The direction ratios of the line are (2, 3, 4). Therefore, the direction vector of the line can be represented as:
\[ \vec{d} = (2, 3, 4) \]

### Step 4: Find the vector from point B to point P
The vector from point B to point P is given by:
\[ \vec{BP} = P - B = (5 - (2\lambda + 1), 9 - (3\lambda + 2), 3 - (4\lambda + 3)) \]
This simplifies to:
\[ \vec{BP} = (4 - 2\lambda, 7 - 3\lambda, -1 - 4\lambda) \]

### Step 5: Set the dot product to zero
For point P to be the image of point B with respect to the line, the vector \(\vec{BP}\) must be perpendicular to the direction vector \(\vec{d}\). Therefore, we set the dot product to zero:
\[ \vec{BP} \cdot \vec{d} = 0 \]
Calculating the dot product:
\[ (4 - 2\lambda) \cdot 2 + (7 - 3\lambda) \cdot 3 + (-1 - 4\lambda) \cdot 4 = 0 \]
Expanding this gives:
\[ 8 - 4\lambda + 21 - 9\lambda - 4 - 16\lambda = 0 \]
Combining like terms:
\[ 25 - 29\lambda = 0 \]

### Step 6: Solve for \(\lambda\)
Now, we solve for \(\lambda\):
\[ 29\lambda = 25 \implies \lambda = \frac{25}{29} \]

### Step 7: Substitute \(\lambda\) back into the parametric equations
Now, we substitute \(\lambda\) back into the parametric equations to find the coordinates of point B:
\[ x_B = 2\left(\frac{25}{29}\right) + 1 = \frac{50}{29} + \frac{29}{29} = \frac{79}{29} \]
\[ y_B = 3\left(\frac{25}{29}\right) + 2 = \frac{75}{29} + \frac{58}{29} = \frac{133}{29} \]
\[ z_B = 4\left(\frac{25}{29}\right) + 3 = \frac{100}{29} + \frac{87}{29} = \frac{187}{29} \]

### Step 8: Find the coordinates of the image point P'
The image point P' can be found using the midpoint formula. The midpoint M of points P and P' is point B. Thus, we have:
\[ M_x = \frac{5 + x_{P'}}{2} = \frac{79}{29} \]
\[ M_y = \frac{9 + y_{P'}}{2} = \frac{133}{29} \]
\[ M_z = \frac{3 + z_{P'}}{2} = \frac{187}{29} \]

Solving for \(x_{P'}\), \(y_{P'}\), and \(z_{P'}\):
1. For \(x_{P'}\):
\[ \frac{5 + x_{P'}}{2} = \frac{79}{29} \implies 5 + x_{P'} = \frac{158}{29} \implies x_{P'} = \frac{158}{29} - 5 = \frac{158 - 145}{29} = \frac{13}{29} \]

2. For \(y_{P'}\):
\[ \frac{9 + y_{P'}}{2} = \frac{133}{29} \implies 9 + y_{P'} = \frac{266}{29} \implies y_{P'} = \frac{266}{29} - 9 = \frac{266 - 261}{29} = \frac{5}{29} \]

3. For \(z_{P'}\):
\[ \frac{3 + z_{P'}}{2} = \frac{187}{29} \implies 3 + z_{P'} = \frac{374}{29} \implies z_{P'} = \frac{374}{29} - 3 = \frac{374 - 87}{29} = \frac{287}{29} \]

### Final Answer
Thus, the image of point P(5, 9, 3) on the line is:
\[ P'\left(\frac{13}{29}, \frac{5}{29}, \frac{287}{29}\right) \]