A cylindrical metallic wire of initial resistance 5 is drawn so...

Let the initial length of the wire be $L_0$ and the initial area of cross-section be $A_0$.

The initial resistance is given by $R_0 = \rho \dfrac{L_0}{A_0} = 5 \Omega$.

When the wire is drawn to three times its original length, the new length is $L = 3L_0$. Since the volume of the wire remains constant, $A_0 L_0 = A L$, which gives the new area of cross-section as $A = \dfrac{A_0}{3}$.

The new resistance of the elongated wire is $R = \rho \dfrac{L}{A} = \rho \dfrac{3L_0}{\dfrac{A_0}{3}} = 9 \rho \dfrac{L_0}{A_0} = 9R_0$.

Substituting $R_0 = 5 \Omega$, we get $R = 9 \times 5 = 45 \Omega$.

The wire is then folded to form a closed equilateral triangle. The total resistance of $45 \Omega$ is divided equally among the three sides of the triangle.

The resistance of each side of the triangle is $R_{side} = \dfrac{45}{3} = 15 \Omega$.

To find the equivalent resistance between any two vertices, we observe that one side of the triangle is in parallel with the series combination of the other two sides.

The resistance of the series combination is $15 \Omega + 15 \Omega = 30 \Omega$.

The equivalent resistance $R_{eq}$ between the two vertices is the parallel combination of $15 \Omega$ and $30 \Omega$:

$R_{eq} = \dfrac{15 \times 30}{15 + 30} = \dfrac{450}{45} = 10 \Omega$.

Answer: $10 \Omega$