If lines `x=y=za n dx=y/2=z/3` and third line passing through `(1,1,1)` form a triangle of area `sqrt(6)` units, then the point of intersection of third line with the second line will be a. `(1,2,3)` b. `2,4,6` c. `4/3,6/3,(12)/3` d. none of these

Options

A. (1,2,3)
B. (2,4,6)
C. `((4)/(3),(8)/(3),(12)/(3))`
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the lines
We have two lines given as:
1. Line L1: \( x = y = z \)
2. Line L2: \( x = \frac{y}{2} = \frac{z}{3} \)

We also have a third line passing through the point \( (1, 1, 1) \).

### Step 2: Parametrize the lines
For Line L1, we can express it in parametric form:
- Let \( x = t \), then \( y = t \) and \( z = t \).
So, the parametric form of Line L1 is:
\[ (t, t, t) \]

For Line L2, we can express it in parametric form:
- Let \( x = s \), then \( y = 2s \) and \( z = 3s \).
So, the parametric form of Line L2 is:
\[ (s, 2s, 3s) \]

### Step 3: Find the direction ratios
The direction ratios for Line L1 (OA) are:
- \( \text{Direction ratios of OA} = (1, 1, 1) \)

The direction ratios for Line L2 (OB) are:
- \( \text{Direction ratios of OB} = (1, 2, 3) \)

### Step 4: Area of the triangle
The area of the triangle formed by points O (origin), A (1, 1, 1), and B (unknown point on Line L2) is given as \( \sqrt{6} \).

The area of a triangle formed by vectors OA and OB can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \| \text{OA} \times \text{OB} \| \]
where \( \| \text{OA} \times \text{OB} \| \) is the magnitude of the cross product of the vectors OA and OB.

### Step 5: Calculate the cross product
Let \( \text{OA} = (1, 1, 1) \) and \( \text{OB} = (s, 2s, 3s) \).

The cross product \( \text{OA} \times \text{OB} \) is calculated as follows:
\[ \text{OA} \times \text{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ s & 2s & 3s \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(1 \cdot 3s - 1 \cdot 2s) - \hat{j}(1 \cdot 3s - 1 \cdot s) + \hat{k}(1 \cdot 2s - 1 \cdot s) \]
\[ = \hat{i}(3s - 2s) - \hat{j}(3s - s) + \hat{k}(2s - s) \]
\[ = \hat{i}(s) - \hat{j}(2s) + \hat{k}(s) \]
So, \( \text{OA} \times \text{OB} = (s, -2s, s) \).

### Step 6: Magnitude of the cross product
The magnitude is given by:
\[ \| \text{OA} \times \text{OB} \| = \sqrt{s^2 + (-2s)^2 + s^2} = \sqrt{s^2 + 4s^2 + s^2} = \sqrt{6s^2} = s\sqrt{6} \]

### Step 7: Area of the triangle
Using the area formula:
\[ \text{Area} = \frac{1}{2} \| \text{OA} \times \text{OB} \| = \frac{1}{2} s\sqrt{6} \]

Setting this equal to the given area \( \sqrt{6} \):
\[ \frac{1}{2} s\sqrt{6} = \sqrt{6} \]
Dividing both sides by \( \sqrt{6} \):
\[ \frac{1}{2} s = 1 \implies s = 2 \]

### Step 8: Find the point of intersection
Substituting \( s = 2 \) into the parametric equations of Line L2:
\[ x = 2, \quad y = 2 \cdot 2 = 4, \quad z = 3 \cdot 2 = 6 \]
Thus, the point of intersection of the third line with the second line is \( (2, 4, 6) \).

### Final Answer
The point of intersection of the third line with the second line is:
\[ \boxed{(2, 4, 6)} \]