$3 \times (32)^{\dfrac{2}{5}} \times 7^0$ is equal to :

1. 0

2. 12

3. 1

4. 9

Simplifying the expression :

$\Rightarrow 3 \times (32)^{\dfrac{2}{5}} \times 7^0 = 3 \times (2^5)^{\dfrac{2}{5}} \times 7^0 \\[1em] = 3 \times 2^2 \times 1 \\[1em] = 3 \times 4 \\[1em] = 12.$

Hence, Option 2 is the correct option.