The value of `lambda` for which the loci arg `z = (pi)/(6)` and `| z-2 sqrt(3)i | = lambda` on the argand plane touch each other is

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the value of \( \lambda \) for which the loci defined by \( \arg z = \frac{\pi}{6} \) and \( |z - 2\sqrt{3}i| = \lambda \) touch each other on the Argand plane.

### Step-by-step Solution:

1. **Understanding the loci**:
- The locus \( \arg z = \frac{\pi}{6} \) represents a straight line in the Argand plane that makes an angle of \( \frac{\pi}{6} \) with the positive x-axis. This line can be expressed as \( y = \tan\left(\frac{\pi}{6}\right)x = \frac{1}{\sqrt{3}}x \).
- The locus \( |z - 2\sqrt{3}i| = \lambda \) represents a circle centered at \( (0, 2\sqrt{3}) \) with radius \( \lambda \).

2. **Finding the center and radius of the circle**:
- The center of the circle is at the point \( (0, 2\sqrt{3}) \) and the radius is \( \lambda \).

3. **Finding the distance from the center to the line**:
- The equation of the line can be rewritten in the form \( \sqrt{3}x - y = 0 \).
- The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \( A = \sqrt{3}, B = -1, C = 0 \) and the center of the circle is \( (0, 2\sqrt{3}) \).

4. **Substituting the values**:
- Substitute \( (x_0, y_0) = (0, 2\sqrt{3}) \):
\[ d = \frac{|\sqrt{3}(0) - (2\sqrt{3}) + 0|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = \frac{| - 2\sqrt{3}|}{\sqrt{3 + 1}} = \frac{2\sqrt{3}}{2} = \sqrt{3} \]

5. **Setting the distance equal to the radius**:
- For the line and the circle to touch, the distance \( d \) must equal the radius \( \lambda \):
\[ \lambda = \sqrt{3} \]

6. **Conclusion**:
- The value of \( \lambda \) for which the loci touch each other is \( \sqrt{3} \).

### Final Answer:
\[ \lambda = \sqrt{3} \]