Intuitive explanation for p ∨ q → r ≡ ( p → r) ∧ (q → r) Although, it is possible to prove the above equivalence using truth tables, I don't know how to prove it without using truth tables.Can someone explain it in plain english?

This is sometimes referred to as Inference by Cases.
It's easy if you bear in mind that $\mathcal{S} \Rightarrow \mathcal{T}$ is equivalent to $(\lnot \mathcal{S}) \lor \mathcal{T}$ where $\mathcal{S}$ and $\mathcal{T}$ are any wffs.
\begin{align}
[(P \lor Q) \Rightarrow R] &\Leftrightarrow [\lnot(P \lor Q) \lor R] \\
&\Leftrightarrow [(\lnot P \land \lnot Q) \lor R] \quad \text{DeMorgan's} \\
&\Leftrightarrow [(\lnot P \lor R) \land (\lnot Q \lor R)] \\
&\Leftrightarrow [(P \Rightarrow R) \land (Q \Rightarrow R)].
\end{align}
