Are you good at manipulating?

Through Vieta you can see that .
$p+q+r=-5\\pq+qr+rp= -4\\pqr=-3\\The~exp.=pq(p+q)+qr(q+r)+rp(r+p)\\\implies~The~exp.=pq(-5-r)+qr(-5-p)+rp(-5-q)\\\implies~The~exp.=-5(pq+qr+rp)-3pqr\\implies~The~exp.=-5*(-4) -3*(-3) = \boxed{29}$

$p^2q+p^2r+q^2p+r^2p+q^2r+r^2q = pq(p+q)+qr(q+r)+rp(r+p)$

$\quad = pq(p+q+r) - pqr + qr(p+q+r) - pqr +rp(p+q+r) - pqr$

$\quad = (p+q+r)(pq+qr+rp) - 3pqr$

Using Vieta's Formulas, we have: $p+q+r = -5$ , $pq+qr+rp=-4$ and $pqr = -3$ .

Therefore,

$p^2q+p^2r+q^2p+r^2p+q^2r+r^2q = (p+q+r)(pq+qr+rp) - 3pqr$

$\quad = (-5)(-4) -3 (-3) = 20 + 9 = \boxed{29}$

Through Vieta you can see that

$q+p+r=-5\\ qr+pq+pr=-4\\ pqr=-3$

Now, through newton sums,

${ p }^{ 3}+{ q }^{ 3 }+{ r }^{ 3 }=-194$

You will see why it is necessary. Let the equation that is needed to be evaluate for this question be $Q$

Now,

${ \left( q+p+r \right) }^{ 3 }=\left( { p }^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 } \right) +3\left( Q \right) +6\left( pqr \right)$

So,

$Q=\frac { { \left( q+p+r \right) }^{ 3 }-6\left( pqr \right) -\left( { p }^{ 2 }+{ q }^{ 2 }+{ r }^{ 2 } \right) }{ 3 }$

Substituting the values above, we get:

$Q=\frac { -125+18+194 }{ 3 } =\boxed{29}$

I came across this problem when trying to solve another problem.

Let the polynomial be $P (x)$ .

Adding $2pqr=-6$ , the expression becomes $(p+q)(q+r)(r+p)=(-5-p)(-5-q)(-5-r)=P (-5)=23$ hence the answer is $23-(-6)=29$ .

Oops, I just brute forced the solution:

$\{p,q,r\}=x\text{/.}\, \text{Solve}\left[x^3+5 x^2-4 x+3=0\right]$

$\left\{\frac{1}{3} \left(-37 \sqrt[3]{\frac{2}{511-3 \sqrt{6501}}}-\sqrt[3]{\frac{1}{2} \left(511-3 \sqrt{6501}\right)}-5\right),\\ \frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(511-3 \sqrt{6501}\right)}-\frac{5}{3}+\frac{37 \left(1-i \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{511-3 \sqrt{6501}}},\\ \frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(511-3 \sqrt{6501}\right)}-\frac{5}{3}+\frac{37 \left(1+i \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{511-3 \sqrt{6501}}}\right\}$

$\text{FullSimplify}\left[p^2 q+p^2 r+p q^2+p r^2+q^2 r+q r^2\right]$

$29$

Using Vieta's Formulas $p+q+r=-5$ $pq+pr+qr=-4$ $-pqr=3$

Then $(p+q+r)(pq+pr+qr)=-5\times-4$ ${ p }^{ 2 }q+{ p }^{ 2 }r+{ q }^{ 2 }p+r^{ 2 }p+{ q }^{ 2 }r+r^{ 2 }q+3pqr=20$ ${ p }^{ 2 }q+{ p }^{ 2 }r+{ q }^{ 2 }p+r^{ 2 }p+{ q }^{ 2 }r+r^{ 2 }q+3pqr-3pqr=20+9$ $\boxed{{ p }^{ 2 }q+{ p }^{ 2 }r+{ q }^{ 2 }p+r^{ 2 }p+{ q }^{ 2 }r+r^{ 2 }q=29}$

The expression is actually equal to $(p+q+r)(pq+pr+qr) - 3pqr$ and by Vieta, $p+q+r= -5\\pq+pr+qr = -4\\pqr = -3$ substituting will give us $-5 \times -4 + (- 3 \times -3) = 20 + 9 = 29$