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Given that,

- $x<0$
- $0<y<1$
- $z>1$

Let’s check all of the options.

$A. \; (x^{2} – z^{2})$ has to be positive.

Let’s take $x=-1, z=2$

Now, $(-1)^{2} – (2)^{2}>0$

$\Rightarrow 1-4>0$

$\Rightarrow \boxed{-3>0 \; (False)}$

$B. \; yz can be less than one.

$yz<1$

$\Rightarrow \frac{1}{4} \times 2<1$

$\Rightarrow \boxed{\frac{1}{2}<1 \; (True)}$

$C: xy$ can never be zero.

$\Rightarrow xy \ne 0$

$\Rightarrow \boxed{xy \ne 0} (True)$

$D. \; (y^{2} – z^{2})$ is always negative.

$\Rightarrow y^{2} – z^{2} <0$

Here $y<z$

$\Rightarrow y^{2} < z^{2}$

$\Rightarrow \boxed{y^{2} - z^{2} < 0 (True)Always}$

Correct Answer $:\text{A}$