Answer
$a_1=4$
Work Step by Step
Let's note by $\{a_n\}$ the given geometric sequence. We have:
$$\begin{cases}
a_1+a_2+a_3=7\\
r=3.
\end{cases}$$
In order to find $a_1$, we use the formula for the sum of the first $n$ terms:
$$S_n=\dfrac{a_1(1-r^n)}{1-r}$$
which we solve for $a_1$:
$$\begin{align*}
S_3&=\dfrac{a_1(1-r^3)}{1-r}\\
52&=\dfrac{a_1(1-3^3)}{1-3}\\
52&=13a_1\\
a_1&=4.
\end{align*}$$
So the first term of the sequence is $4$.
