Answer
When switch $S$ is closed, $V_1$ decreases from $6.0~V$ to $4.0~V$
Work Step by Step
When the switch is open, the current flows through $R_3$ and then $R_1$. The equivalent resistance $R_{13} = 12~\Omega$
Then the current through $R_3$ and $R_1$ is $\frac{12~V}{12~\Omega} = 1.0~A$
We can find the potential difference across $R_1$:
$V_1 = (1.0~A)(6.0~\Omega) = 6.0~V$
When the switch is closed, we can find the equivalent resistance of $R_1$ and $R_2$:
$R_{12} = \frac{1}{6.0~\Omega}+\frac{1}{6.0~\Omega}$
$R_{12} = 3.0~\Omega$
Then the equivalent resistance of the circuit is $9.0~\Omega$
The total current in the circuit is $i = \frac{12~V}{9.0~\Omega} = 1.33~A$
The potential difference across $R_3$ is $V_3 = (1.33~A)(6.0~\Omega) = 8.0~V$
Then the potential difference across $R_1$ is $V_1 = 12~V-8.0~V = 4.0~V$
When switch $S$ is closed, $V_1$ decreases from $6.0~V$ to $4.0~V$
