1. Set up the configuration

Let the two charges $q$ and $3q$ be separated by distance $r$. Assume both charges are of the same sign, so the electric field can be zero only at some point between them.

Let the point where resultant electric field is zero be at distance $x$ from charge $q$. Then its distance from charge $3q$ is:

$r-x$

2. Write the condition for zero resultant electric field

At the null point, magnitudes of the two electric fields must be equal:

$\frac{kq}{x^2} = \frac{k(3q)}{(r-x)^2}$

Cancel $k$ and $q$:

$\frac{1}{x^2} = \frac{3}{(r-x)^2}$

3. Solve the equation

Cross-multiplying,

$(r-x)^2 = 3x^2$

Taking square root,

$r-x = \sqrt{3}\,x$

(We take the positive root since distances are positive and the point lies between the charges.)

So,

$r = x(1+\sqrt{3})$

Hence,

$x = \frac{r}{1+\sqrt{3}}$

4. Match with the options

This corresponds to:

$\boxed{\frac{r}{1+\sqrt{3}}}$

So the correct option is C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.