Statement-I The point `A(1, 0, 7)` is the mirror image of the point `B(1,6, 3)` in the line `(x)/(1)=(y-1)/(2)=(z-2)/(3)`.
Statement-II The line `(x)/(1)=(y-1)/(2)=(z-2)/(3)` bisect the line segment joining `A(1, 0, 7) and B(1, 6, 3)`.

Options

A. Statement-I is true, Statement-II is also true, Statement-II is not the correct explanation of Statement-I.
B. statement-I is true, Statement-II is false.
C. Statement-I is false, Statement -II is true.
D. statement-I is true, Statement II is also true, Statement-II is the correct explanation of Statement-I.

Answer

To solve the problem, we need to analyze both statements regarding the points A(1, 0, 7) and B(1, 6, 3) in relation to the line given by the equations \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} \).

### Step 1: Find the Midpoint of Segment AB

The midpoint \( P \) of the segment joining points \( A(1, 0, 7) \) and \( B(1, 6, 3) \) can be calculated using the midpoint formula:

\[ P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]

Substituting the coordinates of points A and B:

\[ P = \left( \frac{1 + 1}{2}, \frac{0 + 6}{2}, \frac{7 + 3}{2} \right) = \left( 1, 3, 5 \right) \]

### Step 2: Check if Point P Lies on the Line

Next, we need to check if point \( P(1, 3, 5) \) lies on the line defined by \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} \).

Substituting \( P \) into the line equations:

1. For \( x = 1 \):
\[ \frac{1}{1} = 1 \]

2. For \( y = 3 \):
\[ \frac{3 - 1}{2} = \frac{2}{2} = 1 \]

3. For \( z = 5 \):
\[ \frac{5 - 2}{3} = \frac{3}{3} = 1 \]

Since all three equations yield the same result (1), point \( P \) lies on the line.

### Step 3: Determine Direction Cosines of AB and L

Next, we find the direction cosines of the line segment \( AB \) and the line \( L \).

1. Direction cosines of \( AB \):
\[ A(1, 0, 7) \quad B(1, 6, 3) \]
The direction vector \( \overrightarrow{AB} = B - A = (1-1, 6-0, 3-7) = (0, 6, -4) \).

The direction cosines are given by:
\[ a_1 = 0, \quad b_1 = 6, \quad c_1 = -4 \]

2. Direction cosines of line \( L \):
The line is given by \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} \), which has direction ratios \( (1, 2, 3) \).

Thus, the direction cosines are:
\[ a_2 = 1, \quad b_2 = 2, \quad c_2 = 3 \]

### Step 4: Check Perpendicularity

To check if \( AB \) is perpendicular to \( L \), we compute the dot product of their direction cosines:

\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \cdot 1 + 6 \cdot 2 + (-4) \cdot 3 \]
\[ = 0 + 12 - 12 = 0 \]

Since the dot product is zero, \( AB \) is perpendicular to \( L \).

### Conclusion

- **Statement I**: The point \( A(1, 0, 7) \) is the mirror image of point \( B(1, 6, 3) \) in line \( L \) is **True**.
- **Statement II**: The line \( L \) bisects the line segment joining \( A \) and \( B \) is also **True**.

However, Statement II does not correctly explain Statement I, as the midpoint condition alone does not imply that \( A \) is the mirror image of \( B \) in line \( L \).