A uniform metallic wire has a resistance of 18 $$\Omega$$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is:

We have a uniform metallic wire whose total resistance is given as $$18\,\Omega$$. Because a metallic wire is uniform, its resistance is directly proportional to its length, as stated by the basic relation $$R=\rho\frac{L}{A},$$ where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area. Since $$\rho$$ and $$A$$ remain the same for the whole wire, halving or thirding the length simply halves or thirds the resistance.

Now the wire is bent to form an equilateral triangle. An equilateral triangle has three sides of equal length, so the original length of the wire is divided equally into three parts. Therefore each side uses exactly one-third of the original length. Because resistance is proportional to length, each side acquires one-third of the original resistance:

$$R_{\text{side}}=\frac{1}{3}\times 18\,\Omega = 6\,\Omega.$$

Label the three vertices as $$A,\,B,\,C$$. Each of the three sides $$AB,\,BC,\,CA$$ now has a resistance of $$6\,\Omega$$.

The question asks for the resistance measured between any two vertices; let us choose vertices $$A$$ and $$B$$. There are two distinct electrical paths connecting these vertices:

1. The direct side $$AB$$ itself, whose resistance is $$6\,\Omega$$.

2. The indirect path that goes from $$A$$ to $$C$$ and then from $$C$$ to $$B$$, passing through two sides in series. The resistance of this longer path is

$$R_{AC}+R_{CB}=6\,\Omega+6\,\Omega=12\,\Omega.$$

Thus, between vertices $$A$$ and $$B$$ we effectively have two resistances—one of $$6\,\Omega$$ and the other of $$12\,\Omega$$—connected in parallel. For two resistors $$R_1$$ and $$R_2$$ in parallel, the equivalent resistance $$R_{\text{eq}}$$ is given by the formula

$$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}.$$

Substituting $$R_1=6\,\Omega$$ and $$R_2=12\,\Omega$$ we get

$$\frac{1}{R_{\text{eq}}}=\frac{1}{6}+\frac{1}{12} =\frac{2}{12}+\frac{1}{12} =\frac{3}{12} =\frac{1}{4}.$$

Taking the reciprocal on both sides,

$$R_{\text{eq}}=4\,\Omega.$$

This equivalent resistance is the same no matter which pair of vertices is chosen, because all three sides are identical.

Hence, the correct answer is Option D.