Principle of the analogy
Thermal conductance, $\displaystyle{\frac{KA}{L}}$, is the amount of heat that passes through a slab of area ($A$) and thickness ($L$) in unit time, when the opposite faces of the surface differ in tmperature by $1K$. It's reciprocal is the thermal resistance $R_T$,
$$R_T=\frac{L}{KA}\longrightarrow (1)$$
Hence, the rate of heat transfer from one face to the other is given by:
$$\frac{Q}{t}=\frac{KA\Delta T}{L}\longrightarrow(2)$$
which follows from the definition of thermal conductance, when the opposite faces are having a difference in temperature by $\Delta T$. From equation $(1)$, we have
$$\frac{Q}{t}=\frac{\Delta T}{R_T}\longrightarrow(3)$$
Let, the heat current (amount of heat transferred in unit time) be denoted by $I_H$. Hence
$$I_H=\frac{Q}{t}=\frac{\Delta T}{R_T}\longrightarrow(4)$$
Now, comparing this equation with the Ohm's law, we have the electric current, which is the amount of charge flowing per second
$$I=\frac{V}{R}\longrightarrow(5)$$
Comparing both equations, we can make an analogy. We have in equation $(4)$, $\Delta T$, the difference in temperature in analogy with the electric potential $V$, and $R_T$, the thermal resistance in analogy with the electrical resistance $R$.
Just as like a voltage (or more precisely, potential difference) makes a current flow through a metallic conductor, the temperature difference between the two ends of a metallic rod makes the heat flow to happen from a point of higher temperature to that of lower temperature. This makes sense. Hence this analogy is useful.
Hence it can be extended to Kirchhoff's law in electricity also, as the only difference in the two cases are that in one heat is the energy transferred, while in the other, electrical energy is being transferred.
How to work out the problem using this analogy?
One of the ends of the three rods start from a common point, say a heat source maintained at a constant temperature $T$. The direction of current flow states that $T$ is the high temperature region.
The current starts from the junction. Hence at the junction, by the conservation of energy, the currents should be added to zero, just as in Kirchhoff's rule. How's that? Simple: For current flow to take place, there should be temperature difference between the two points. If you take the junction point alone, which is at a single temperature, there is no current flow. Hence the currents add up to zero at the junction. But it flows to the other ends, as obviously there is some temperature difference.
First, use the principle that at the junction $T$,
$$i_{Al}+i_{S}+i_{Cu}=0\longrightarrow(6)$$
If the temperature at the junction is $T$ and that at the end points of the rods are $T_{Al}$, $T_{S}$, and $T_{Cu}$, then the corresponding temperature differences become $(\Delta T)_{Al}=(T-T_{Al})$; $(\Delta T)_{S}=(T-T_{S})$; $(\Delta T)_{Cu}=(T-T_{Cu})$. Now using equation $(4)$, $(6)$ reads
$$\frac{(T-T_{Al})}{R_{Al}}+\frac{(T-T_{S})}{R_{S}}+\frac{(T-T_{Cu})}{R_{Cu}}=0$$
$$T\left[\frac{1}{R_{Al}}+\frac{1}{R_{S}}+\frac{1}{R_{Cu}}\right]=\frac{T_{Al}}{R_{Al}}+\frac{T_{S}}{R_{S}}+\frac{T_{Cu}}{R_{Cu}}\longrightarrow(7)$$.
Now, knowing the resistances, we can calculate $T$.
