A line `L_1` passing through a point with position vector `p=i+2j+3k` and parallel `a=i+2j+3k`, Another line `L_2` passing through a point with position vector to `b=3i+j+2k`. and parallel to b=3i+j+2k.
Q. Equation of a line passing through the point `(2, -3, 2)` and equally inclined to the line `L_1 and L_2` may equal to

Options

A. a. `(x-2)/(2)=(y-3)/(-1), (z-2)/(1)`
B. b. `(x-2)/(2)=y+3=z-2`
C. c. `(x-2)/(-4)=(y+3)/(3), (z-5)/(2)`
D. d. `(x+2)/(4)=(y+3)/(3), (z-2)/(-5)`

Answer

To solve the problem, we need to find the equation of a line that passes through the point \( (2, -3, 2) \) and is equally inclined to the lines \( L_1 \) and \( L_2 \).

### Step 1: Identify the direction vectors of the lines \( L_1 \) and \( L_2 \)

The line \( L_1 \) is given to be parallel to the vector \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \).

The line \( L_2 \) is given to be parallel to the vector \( \mathbf{b} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k} \).

### Step 2: Find the direction vectors of the angle bisectors

To find the direction vector of the line that is equally inclined to both \( L_1 \) and \( L_2 \), we need to find the angle bisectors of the vectors \( \mathbf{a} \) and \( \mathbf{b} \).

The unit vectors in the direction of \( \mathbf{a} \) and \( \mathbf{b} \) are:

\[ \hat{a} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}}{\sqrt{14}} \]

\[ \hat{b} = \frac{\mathbf{b}}{|\mathbf{b}|} = \frac{3\mathbf{i} + \mathbf{j} + 2\mathbf{k}}{\sqrt{3^2 + 1^2 + 2^2}} = \frac{3\mathbf{i} + \mathbf{j} + 2\mathbf{k}}{\sqrt{14}} \]

### Step 3: Find the direction vector of the angle bisector

The direction vector of the angle bisector can be expressed as:

\[ \mathbf{d_1} = \hat{a} + \hat{b} = \frac{\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}}{\sqrt{14}} + \frac{3\mathbf{i} + \mathbf{j} + 2\mathbf{k}}{\sqrt{14}} = \frac{(1+3)\mathbf{i} + (2+1)\mathbf{j} + (3+2)\mathbf{k}}{\sqrt{14}} = \frac{4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}}{\sqrt{14}} \]

### Step 4: Write the equation of the line

The line we are looking for passes through the point \( (2, -3, 2) \) and has the direction vector \( \mathbf{d_1} = 4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k} \).

The parametric equations of the line can be written as:

\[ \begin{align*} x &= 2 + 4t \\ y &= -3 + 3t \\ z &= 2 + 5t \end{align*} \]

### Step 5: Write the symmetric form of the line

The symmetric form of the line can be expressed as:

\[ \frac{x - 2}{4} = \frac{y + 3}{3} = \frac{z - 2}{5} \]

### Final Answer

Thus, the equation of the line passing through the point \( (2, -3, 2) \) and equally inclined to the lines \( L_1 \) and \( L_2 \) is:

\[ \frac{x - 2}{4} = \frac{y + 3}{3} = \frac{z - 2}{5} \]