Solution:

First,

 Find the prime factors of the given integers: \[84,\text{ }90\text{ }and\text{ }120\]     

For,

\[\begin{array}{*{35}{l}}

   ~84\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }7  \\

   90\text{ }=\text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }5  \\

   120\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }5  \\

\end{array}\]

Now,

\[\begin{array}{*{35}{l}}

   L.C.M\text{ }of\text{ }84,\text{ }90\text{ }and\text{ }120\text{ }=\text{ }{{2}^{3}}~\times \text{ }{{3}^{2}}~\times \text{ }5\text{ }\times \text{ }7  \\

   \therefore L.C.M\text{ }\left( 84,\text{ }90,\text{ }120 \right)\text{ }=\text{ }2520  \\

\end{array}\]

And,\[\begin{array}{*{35}{l}}

   ~  \\

   ~H.C.F\text{ }\left( 84,\text{ }90\text{ }and\text{ }120 \right)\text{ }=\text{ }6  \\

\end{array}\]

Solution:

First,

find the prime factors of the given integers: \[24,\text{ }15\text{ }and\text{ }36\]         

For,

\[\begin{array}{*{35}{l}}

   24\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }x\text{ }2\text{ }x~3  \\

   15\text{ }=\text{ }3\text{ }\times \text{ }5  \\

   36\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3  \\

\end{array}\]

Now,

\[LCM\text{ }of\text{ }24,\text{ }15\text{ }and\text{ }36\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }5\text{ }=\text{ }{{2}^{3}}~x\text{ }{{3}^{2}}~x\text{ }5\]

∴\[LCM\text{ }\left( 24,\text{ }15,\text{ }36 \right)\text{ }=\text{ }360\]

And,    

\[HCF\text{ }\left( 24,\text{ }15\text{ }and\text{ }36 \right)\text{ }=\text{ }3\]