First, we need to find the equivalent resistance. We can find it as follows:
[tex]\begin{gathered} \frac{1}{Req}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3} \\ so: \\ \frac{1}{Req}=\frac{1}{15}+\frac{1}{5}+\frac{1}{12} \\ \frac{1}{Req}=\frac{7}{20} \\ Req=\frac{20}{7} \\ Req=2.857\Omega \end{gathered}[/tex]Now, we can find the voltage using Ohm's law:
[tex]\begin{gathered} V=IR \\ so: \\ V=10.5\cdot2.857 \\ V=30V \end{gathered}[/tex]Now, since the circuit is in parallel, the voltage is the same for every resistance. So, let's find the current for each resistance using Ohm's law:
For the resistance of 15:
[tex]I_{15}=\frac{V}{R}=\frac{30}{15}=2A[/tex]For the resistance of 5:
[tex]I_5=\frac{V}{R}=\frac{30}{5}=6A[/tex]For the resistance of 12:
[tex]I_{12}=\frac{V}{R}=\frac{30}{12}=2.5A[/tex]Completing the table using the previous data:
