The total number of integral solutions for x, y, z such that $\mathrm{xyz}=24$, is

$$\begin{gathered} xyz=2^3\left(3\right) \\ \mathrm{No} \mathrm{of} \mathrm{positive} \mathrm{solutions} =\left({ }^{3+3-1}{\mathrm{C}}_{3-1}\right)\left({ }^{1+3-1}{\mathrm{C}}_{3-1}\right) \\ {=}^5{\mathrm{C}}_2\left({ }^3{\mathrm{C}}_2\right)=10\left(3\right)=30 \end{gathered}$$

Since negative values of  x,y,z is also allowed but since product is positive and hence any two of them may be negative

Number of negative integral solutions=${ }^3{\mathrm{C}}_2\left(30\right)=90$

Hence total number of integral solutions=30+90=120