Domain of truth

$\color{#69047E} {\boldsymbol {log(xyz) = log(x+y+z)}}$ log x is only defined when x>0 . So, we can rewrite the equation as, $\color{#3D99F6}{\boldsymbol {x+y+z=xyz----(i)}}$ Only if, $\color{teal} {\boldsymbol{xyz>0} }$ .

For the $\color{teal} {inequality}$ to be true, there can be two possible cases,

$\color{teal} {\boldsymbol {\text{Case 1: } x, y, z>0}}$

Let's assume, $\boldsymbol {x\leq y\leq z}$ $\boldsymbol {\Rightarrow x+y+z\leq 3z}$ $\boldsymbol{\Rightarrow xyz\leq 3z}$ $\boldsymbol {\Rightarrow xy\leq 3}$ So, we get only 3 possible integer pairs of $\boldsymbol {(x, y):}$ $\boldsymbol {(1,1),(1,2),(1,3)}$ Putting these values in $\color{#3D99F6} {\text{equation (i)}}$ , we get only one possible integer triple, $\boldsymbol {(x, y, z) =(1,2,3)}$ Now, if we take out the condition of $\boldsymbol {x\leq y\leq z}$ , the total number of triples we get is, $\color{#69047E} {\boldsymbol {3!=\boxed{6}}}$

$\color{teal} {\boldsymbol {\text{Case 2: Any 2 of the 3 variables is negative and the other one is positive.}}}$

Let's assume $\boldsymbol {x\leq y <0<z}$ .

Then, $\boldsymbol {x+y<z}$ $\boldsymbol {\Rightarrow xyz-z<z}$ $\boldsymbol {\Rightarrow xyz<2z}$ $\boldsymbol {\Rightarrow xy<2}$

This time we get 2 possible integer pairs of $\boldsymbol {(x, y) :}$ $\boldsymbol {(-1,-1),(-2,-1)}$ But none of the pairs gives a positive integer value of z, when we put them in $\color{#3D99F6} {\text{equation (i)}}$ .

Thus, the total number of ordered triples is $\color{#69047E} {\boxed{6}}$ .

From inspection we find that the identity xyz=x+y+z is only true for positive integers 1, 2 and 3. The identity doesn't hold for any other combination in the given interval. So the total number of ordered triplets is 3!=6