To find the total resistance $R_{\text{total}}$ in this circuit, we need to break down the resistors into series and parallel combinations. Let’s work through it step by step based on the image provided:

Resistor Breakdown:

1. The top branch consists of the resistors $3 \, \Omega$, $6 \, \Omega$, and $2 \, \Omega$ in series.

   • Series Rule: Resistors in series add directly.
   • So, the total resistance of the top branch is: $R_{\text{top}} = 3 \, \Omega + 6 \, \Omega + 2 \, \Omega = 11 \, \Omega$

2. The bottom branch consists of the resistors $4 \, \Omega$, $12 \, \Omega$, and $10 \, \Omega$, where $4 \, \Omega$ and $12 \, \Omega$ are in series and their combination is in parallel with the $10 \, \Omega$ resistor.

   • First, combine $4 \, \Omega$ and $12 \, \Omega$ (series): $R_{\text{bottom-series}} = 4 \, \Omega + 12 \, \Omega = 16 \, \Omega$
   • Now, combine $16 \, \Omega$ and $10 \, \Omega$ in parallel: $\frac{1}{R_{\text{bottom}}} = \frac{1}{16 \, \Omega} + \frac{1}{10 \, \Omega}$ Let's calculate this explicitly.

I'll now calculate the total resistance step by step.The total resistance of the bottom branch, after combining $4 \, \Omega$, $12 \, \Omega$, and $10 \, \Omega$, is approximately $6.15 \, \Omega$.

Next Steps:

Now, we have two resistances: