Determine two consecutive multiples

Let the required number be $3 x$ and $(3 x+3)$

Then according to question

$(3 x)(3 x+3)=270$

$9 x^{2}+9 x-270=0$

$9\left(x^{2}+x-30\right)=0$

$x^{2}+x-30=0$

$x^{2}+x-30=0$

$x^{2}-5 x+6 x-30=0$

$x(x-5)+6(x-5)=0$

$(x-5)(x+6)=0$

$(x-5)=0$

$x=5$

Or

$(x+6)=0$

$x=-6$

Since, x being a positive number, so x cannot be negative.

Therefore,

When $x=5$ then positive integer

$3 x=3 \times 5$

$=15$

And

$3 x+3=3 \times 5+3$

$=18$

Thus, three consecutive positive integer be 15,18