If l^x = m^y = n^z and lmn = 1, then yz + zx + xy =

1. 0

2. 1

3. -1

4. $\dfrac{1}{2}$

Let us consider l^x = m^y = n^z = k, and lmn = 1

From, l^x = k, we get $l = k^{\dfrac{1}{x}}$

From, m^y = k, we get $m = k^{\dfrac{1}{y}}$

From, n^z = k, we get $n = k^{\dfrac{1}{z}}$

Substituting in lmn = 1:

$\Rightarrow k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}} \times k^{\dfrac{1}{z}} = 1 \\[1em] \Rightarrow k^{\left(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\right)} = 1 \\[1em] \Rightarrow k^{\left(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\right)} = k^0$

Equating the exponents:

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

Multiply both sides by xyz:

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0 \\[1em] \Rightarrow xyz \Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big) = 0 \times xyz \\[1em] \Rightarrow \Big(\dfrac{xyz}{x} + \dfrac{xyz}{y} + \dfrac{xyz}{z}\Big) = 0 \\[1em] \Rightarrow yz + zx + xy = 0.$

Hence, option 1 is the correct option.