The official meaning of the assertion that
$$1+r+r^2+r^3+\cdots=S$$
is that
$$\lim_{n \to \infty}(1+r+r^2+\cdots +r^n) =S$$.
By the usual formula for the sum of a finite geometric series,
$$1+r+r^2+\cdots +r^n=\frac{1-r^{n+1}}{1-r}$$
(unless, of course, $r=1$).
It is reasonably clear that if $|r|<1$,
$$\lim_{n\to\infty}\frac{1-r^{n+1}}{1-r}=\frac{1}{1-r}$$
and that if $|r|\ge 1$, then
$$\lim_{n \to \infty}(1+r+r^2+\cdots +r^n)$$
does not exist.
But that is not really your question. What I think you are saying is roughly this: let
$$S=1+r+r^2+r^3+\cdots$$
Then at the level of formal "algebraic" manipulation, we have (??)
$$S-1=rS$$
and "therefore" (??)
$S=1/(1-r)$
(at least if $r \ne 1$).
But do "formal" manipulations always yield correct results? Not necessarily. For example, you have probably already seen plausible-looking algebraic manipulations that appear to yield the absurd "result" that $0=1$ (usually here the flaw is a carefully hidden division by $0$.)
The formal manipulation yields, if we take $r=1/2$, the perfectly correct result that
$$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots =2$$
This makes physical sense. If you look at the real number line, and put a dot at $1$, then at $1+1/2$, then at $1+1/2+1/4$, and so on, your dots are clearly approaching $2$, and are after a while indistinguishably close to $2$.
Now look at the formal sum
$$S=1+2+2^2+2^3+\cdots$$
Purely formal manipulation then yields that $S=-1$. Does this make any kind of physical sense? Certainly if you add more and more terms of the above series, you are not getting in any sense close to $-1$. And at the crude informal level, it is clear that the sum is "infinite," whatever that may mean. Formal manipulations on non-existent objects can yield absurd results.
So in this case formal manipulation has produced a result that makes no physical sense. There are a number of instances where blind manipulation yields wrong results, so early in university, students are trained to apply
algebraic "rules" only in situations where these rules are valid.
However, the situation is more complicated than that! Once upon a time complex numbers, when they came up in a formal calculation, were dismissed as "absurd." Now they are part of the essential toolkit of the electrical engineer! There are similar phenomena with series.
In combinatorics, for example, formal power series are successfully used, at the purely manipulational level, even when the series technically do not converge. And in various areas of mathematics, divergent series of a suitable type can be very useful, even in computations!
But at this stage, you should be disciplined, and apply formal manipulations only in situations where you know, or have been assured that, they are "safe." Later, perhaps, you can explore situations where venturing beyond safe confines yields interesting and useful results.
The general term of a geometric series is $a r^{n-1}$, where $r$ is the ratio, and $a$ is the first term. In your case, $a=\sqrt{2}$ and $r=\sqrt{3}$. The 12th term is then
$$\sqrt{2} (\sqrt{3})^{11} = 243 \sqrt{6}$$
The sum of the first 12 terms is
$$a \sum_{k=0}^{11} r^k = a \frac{r^{12}-1}{r-1} = \sqrt{2} \frac{728}{\sqrt{3}-1} = 364 (\sqrt{2}+\sqrt{6})$$
Note that I used $3^5 = 243$ and $3^6=729$ in the above.
EDIT
Note that
$$ \sqrt{2} \frac{728}{\sqrt{3}-1} = \sqrt{2} \frac{728}{\sqrt{3}-1} \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{2} (\sqrt{3}+1) 728}{(\sqrt{3})^2-1^2} = \frac{(\sqrt{6}+\sqrt{2}) 728}{3-1}$$
Best Answer
Hint: If you take the ratio $c$ of the 3rd term divided by the 2nd term, you will get
$$(1 + r + r^2) = (1+r)c$$
which gives you a quadratic in $r$ that you can solve. Once you know $r$, then you can solve for $a$ using the 2nd term.