The best way to approach this problem is by substitution.

Quantity A is \(\frac{(x^2(y+z)+y^2(x+z)+z^2(x+y))}{(xyz)}\)

or \(x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{x}+\frac{1}{z})+z(\frac{1}{x}+\frac{1}{y})\)

or \(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}\)

rearrange

\(\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{z}{y}+\frac{y}{z}\).

now Let \(\frac{x}{y}\) = n. Then the first two terms become \(n + \frac{1}{n}\). The minimum value of \(n + \frac{1}{n}\) is 2 when n =1. Or and n can be one if x=y.

Hence a case x=y=z= 5( or any other real positive number) is not possible. Hence Quantity A is always greater

for x=1 y=1 z=2

\(\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{2}{1}+\frac{2}{1}+\frac{1}{2}\)

\(1+1+0.5+2+2+0.5\) =\(7\). Now Quantity A is 7.

Hence A is the correct option.