Let p, q, r $ϵR^{+}$ and $27pqr$ $\ge (p+q+r{)}^3$ and $3p+4q+5r=12$ then $p^3+q^4+r^5$ is equal to

Since $p,q,rϵR^{+}$ , hence we can apply the concept of 
$AM\ge GM$
$\frac{p+q+r}{3}\ge (pqr{)}^{\frac{1}{3}}$

$\frac{(p+q+r{)}^3}{27}\ge pqr$
Or 
$(p+q+r{)}^3\ge 27pqr$. ...(i)
But it is given that $27pqr\ge (p+q+r{)}^3$ ...(ii)
Hence from (i) and (ii), we get 
$(p+q+r{)}^3=27pqr$
Then $p=q=r=1$.
Hence $p^3+q^4+r^5$
$=1+1+1$
$=3$.