Need help to simplify- logic algebra

Question

{( ∼p ∨ ∼q ∨ r) ∧ (p ∧ r)} ∨ {p ∧ (∼q ∨ r)}

I started like that and then i have no idea what to do....can anyone give a hint please.

{( ∼p ∨ ∼q ∨ r) ∧ (p ∧ r)} ∨ (p ∧ ∼q) ∨ (p ∧ r)

Book answer gives only a True Value

Also i cant reach the answer for this one as well

∼p ∧ {∼q ∧ (∼p ∨ q)}

∼p ∧ {(∼q ∧ ∼p) V (∼q ∧ q)

∼p ∧ {(∼q ∧ ∼p) V F)

(∼p ∧ (∼q ∧ ∼p)) V (∼p ∧ F)

(∼p ∧ (∼q ∧ ∼p)) V F

(∼q ∧ ∼p) V F

The book answer gives only a False Value

p.s this is not homework, im just trying to run through a discrete structure book before joining uni for a comp science course soon.

Answer 1

$\{(\lnot p \lor \lnot q \lor r) \land (p \land r)\} \lor \{p \land (\lnot q \lor r)\} \\\{(\lnot p \lor \lnot q \lor r) \land (p \land r)\} \lor \{(p \land \lnot q) \lor(p\land r)\}$

Just use commutation and association and you are almost done.

$~~~[\{(\lnot p \lor \lnot q \lor r) \land (p \land r)\}\lor(p\land r)\}] \lor (p \land \lnot q) $

Keep going... (here's a hint if you want...)

PS: Recall the rule of absorption: $(\psi\land \phi)\lor \phi = \phi$

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Book answer gives only a True Value

Also, the book gives the wrong answer.

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$\lnot p \land \{\lnot q \land (\lnot p \lor q)\} \\ \lnot p \land \{(\lnot q \land \lnot p) \lor (\lnot q \land q)\} \\ \lnot p \land \{(\lnot q \land \lnot p) \lor \mathrm F\}$

Up to here is quite okay. Next, recall that $\phi\lor\mathrm F=\phi$ .

$~~\lnot p \land (\lnot q \land \lnot p)$

What you must do next should be clear.

The book answer gives only a False Value

Also, the wrong answer.