Find two consecutive multiples of 3 whose product is 648.

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,

$3 x \times 3(x+1)=648$

$\Rightarrow 9\left(x^{2}+x\right)=648$

$\Rightarrow x^{2}+x=72$

$\Rightarrow x^{2}+x-72=0$

$\Rightarrow x^{2}+9 x-8 x-72=0$

$\Rightarrow x(x+9)-8(x+9)=0$

$\Rightarrow(x+9)(x-8)=0$

$\Rightarrow x+9=0$ or $x-8=0$

$\Rightarrow x=-9$ or $x=8$

∴ x = 8           (Neglecting the negative value)

When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27

Hence, the required multiples are 24 and 27.