Electronic – Why not all the diodes are in “on state” in the given picture

Your options 2. and 3. are not valid, because the voltage across the OFF diode is not negative.

For your ideal diode to be OFF it must have a negative voltage and zero current, while it should have a positive current and zero voltage when it is ON. Always check for these 2 conditions.

You are stating that your ideal diode has zero current and zero voltage, which is none of its 2 available states when you are analyzing your circuit.

(Please note that this is being theoretically strict (as we should be when analyzing a circuit of ideal components). Practically your diode can have V = 0, I = 0, but it is not to much use when you are analyzing your circuit)

A "better" way involves changing the model of the diode you're using. The ideal diode model is a discontinuous model so the intermediate value theorem doesn't apply. A better model of the diode is the Shockley diode model, which approximates the V-I relationship as an exponential. Unfortunately, this often leads to a non-linear system of equations with no analytical solution so a numerical algorithm is required. One such algorithm is the Newton-Raphson method, which frames a non-linear smooth system of equations as a root-finding problem, then iterates until a suitably close solution is found. Other methods such as bisection may not require smoothness, but only continuity at the cost of slower convergence/other tradeoffs.

For discontinuous models, the only real solution is to guess and check. You might have some intuition built up from practice on what are good guesses, but at the end of the day there's no alternative.

This method can show uniqueness if you check all possible combinations. However, in most practical cases you can analyze by hand it is often assumed that there is a single unique solution because the cases where there isn't a unique solution are ill-posed mathematically, such as this case:

^{simulate this circuit – Schematic created using CircuitLab}

Assuming \$V_1 \gg V_D\$ and the two diodes have identical forward voltage drops (close is not good enough!), D1 and D2 would both be forward biased leading to two parallel voltage sources. It's impossible without additional information to solve this network.