We verify b) and c) (De Morgan's laws) using a) (double-negation law).
a) $\lnot (\lnot P) \leftrightarrow P$.
b) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q)$
then use c) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \land \lnot Q)$ [ rewrite it as : $\lnot [\lnot (\lnot P) \lor \lnot (\lnot Q) ]$ ; now it is of the "form" : $\lnot [\lnot P_1 \lor \lnot Q_1]$; then you must replace $\lnot P_1 \lor \lnot Q_1$ with $\lnot (P_1 \land Q_1)$, by c), that is really : $\lnot (\lnot P \land \lnot Q)$]. In this way you will get :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \land \lnot Q)$
then apply again double-negation to the right-hand side ("cancelling" $\lnot \lnot$) and you will have :
$\lnot (P \lor Q) \leftrightarrow (\lnot P \land \lnot Q)$.
c) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q)$
then use b) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \lor \lnot Q)$ getting :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \lor \lnot Q)$
then apply again double-negation and it's done.
The Concensus Theorem says:
$XY \lor Y'Z \lor XZ = XY \lor XZ$
So your exercise is really a particular application of this!
But let's assume the Concensus Theorem is not available for you to use.
Then you can do:
$(A \land B) \lor (\neg A \land C) \lor (B \land C) =$ (Adjacency)
$(A \land B) \lor (\neg A \land C) \lor (A \land B \land C) \lor (\neg A \land B \land C)=$ (Absorption x 2)
$(A \land B) \lor (\neg A \land C)$
This assumes:
Adjacency
$P = (P \land Q) \lor (P \land \neg Q)$
Absorption
$P = P \lor (P \land Q)$
Best Answer
If $A\land B$ is true, then one of $A\land C$ or $B\land\neg C$ will also be true already, depending on the truth value of $C$.
So adding an $A\land B$ disjunct will not make the expression more true than it already was.
You can show this symbolically by $$ \begin{align} & (B\land \neg C) \lor (A\land B)\lor (A\land C) \\ ={}& (B\land \neg C) \lor [(A\land B)\land (\neg C\lor C)]\lor (A\land C) \\ ={}& (B\land \neg C) \lor (A\land B\land \neg C) \lor (A\land B\land C) \lor (A\land C) \\ ={}& [(B\land \neg C)\land(1\lor A)] \lor [(A\land C)\land(1\lor B)] \\ ={}& [(B\land \neg C)\land1] \lor [(A\land C)\land1] \\ ={}& (B\land \neg C)\lor (A\land C) \end{align}$$