We know $\(3 x+2 y+z=42\)$, where $\(x, y \& z\)$ are positive integers; we need to compare the value of $\(x+y+z\)$ with 18 .

If we put $x=y=z$ in the equation $\(3 x+2 y+z=42\)$, we get $\(6 x=42 \Rightarrow x=7\)$, so the value of $\(x+y+z=21\)$ which implies column $\(A\)$ has higher quantity.


But if we take only $\(y=z\)$, we get $\(3 x+3 y=42 \Rightarrow x+y=14\)$ which is true for many pairs out of which $\((x, y)=(10,4)\)$ is one. So, we get $\(x+y+z=10+4+4=18\)$ which implies column $\(A\)$ quantity equals column B quantity, so the answer is (C)

Since there are many values possible for the sum of $\(\mathrm{x}, \mathrm{y}\)$ \& \(z \) which may or may not be greater than 18 , a unique comparison cannot be made.

Hence the answer is (D).