If [tex]a[/tex] is the first term in the series and [tex]r[/tex] is the common ratio between consecutive terms, then
[tex]a = 3-x[/tex]
[tex]ar = 6 \implies r = \dfrac6{3-x}[/tex]
[tex]ar^2 = 7-5x \implies r^2 = \dfrac{7-5x}{3-x}[/tex]
Solve for [tex]x[/tex].
[tex]\left(\dfrac6{3-x}\right)^2 = \dfrac{7-5x}{3-x}[/tex]
[tex]\dfrac{36}{(3-x)^2} = \dfrac{(7-5x)^2}{(3-x)^2}[/tex]
[tex]36 = (7 - 5x)^2[/tex]
[tex]7 - 5x = \pm 6[/tex]
[tex]-5x = -1 \text{ or } -5x = -13[/tex]
[tex]\boxed{x = \dfrac15 \text{ or } x = \dfrac{13}5}[/tex]
Then the common ratio is one of
[tex]r = \dfrac6{3 - \frac15} \text{ or } r = \dfrac6{3-\frac{13}5}[/tex]
[tex]\boxed{r = \dfrac{15}7 \text{ or } r = 15}[/tex]
and the sum of the first three terms of this series is
[tex]S_3 = a + ar + ar^2 = \dfrac{a(1-r^3)}{1-r}[/tex]
so that
[tex]r = \dfrac{15}7 \implies \boxed{S_3 = \dfrac{758}{35}}[/tex]
or
[tex]r = 15 \implies \boxed{S_3 = \dfrac{482}5}[/tex]
