Three particles starts from origin at the same time with a velocity `2 ms^-1` along positive x-axis, the second with a velocity `6 ms^-1` along negative y-axis, Find the velocity of the third particle along `x = y` line so that the three particles may always lie in a straight line.

Options

A. `-3sqrt3`
B. `3sqrt2`
C. `-3sqrt2`
D. `2sqrt2`

Explanation

To solve the problem, we need to find the velocity of the third particle (C) such that all three particles (A, B, and C) lie on a straight line at all times.

### Step-by-Step Solution:

1. **Identify the velocities of particles A and B**:
- Particle A moves with a velocity of \(2 \, \text{m/s}\) along the positive x-axis.
- Particle B moves with a velocity of \(6 \, \text{m/s}\) along the negative y-axis.

2. **Determine the position of particles A and B after time \(t\)**:
- The position of particle A after time \(t\) is:
\[ \text{Position of A} = (2t, 0) \]
- The position of particle B after time \(t\) is:
\[ \text{Position of B} = (0, -6t) \]

3. **Assume the velocity of particle C**:
- Let the velocity of particle C be \(V\) along the line \(x = y\). Since it moves at a \(45^\circ\) angle, we can express its velocity components as:
\[ V_x = \frac{V}{\sqrt{2}}, \quad V_y = \frac{V}{\sqrt{2}} \]

4. **Determine the position of particle C after time \(t\)**:
- The position of particle C after time \(t\) is:
\[ \text{Position of C} = \left(\frac{Vt}{\sqrt{2}}, \frac{Vt}{\sqrt{2}}\right) \]

5. **Set up the condition for collinearity**:
- For particles A, B, and C to be collinear, the slopes of the lines AB and AC must be equal.
- The slope of line AB is given by:
\[ \text{slope of AB} = \frac{0 - (-6t)}{2t - 0} = \frac{6t}{2t} = 3 \]
- The slope of line AC is given by:
\[ \text{slope of AC} = \frac{\frac{Vt}{\sqrt{2}} - 0}{\frac{Vt}{\sqrt{2}} - 2t} = \frac{V}{\sqrt{2}} \cdot \frac{1}{\frac{V}{\sqrt{2}} - 2} \]

6. **Equate the slopes**:
- Set the slopes equal to each other:
\[ 3 = \frac{V/\sqrt{2}}{(V/\sqrt{2}) - 2} \]

7. **Cross-multiply and solve for \(V\)**:
- Cross-multiplying gives:
\[ 3\left(V/\sqrt{2} - 2\right) = V/\sqrt{2} \]
- Simplifying:
\[ 3V/\sqrt{2} - 6 = V/\sqrt{2} \]
- Rearranging:
\[ 3V/\sqrt{2} - V/\sqrt{2} = 6 \]
\[ 2V/\sqrt{2} = 6 \]
\[ V = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2} \]

8. **Conclusion**:
- The velocity of the third particle C is:
\[ V = 3\sqrt{2} \, \text{m/s} \]