If the number of ordered triplets (x,y,z) of positive integers such that L.C.M(x,y)=3375,L.C.M(y,z)=1125,L.C.M(z,x)=3375 is equal to ‘K’ then k-47 is equal to
 

$\begin{array}{l}3375=5^3\cdot 3^3,1125=5^3\cdot 3^2\\ \text{ Clearly, }{3}^3{{\text{ is a factor of x, and 3}}^2}_{ }\text{ is factor of atleast one of }y\mathrm{\&}z\text{. This can be }\end{array}$

$\text{ done in }5\text{ ways. }$   $\left(let 3^a is factor of y and 3^b is factor ofz then \left(a,b\right)=\left(0,2\right)or \left(1,2\right)or\left(2,2\right)or\left(2,0\right)or\left(2,1\right) \right)$

${\text{ Also,5}}^3\text{ is a factor of atleast two of the numbers }x,y,z\text{ which can be done in }$

$\begin{array}{l}{ }^3{C}_2\times 4-2=10\\ \therefore k=50\end{array}$$( let 5^a is factor of x,5^b is factor of y,5^c is factor of z,$

$then possible triplets of \left(a,b,c\right)=\left(0,3,3\right),\left(1,3,3\right),\left(2,3,3\right),\left(3,3,3\right),\left(3,0,3\right)\left(3,1,3\right)\left(3,2,3\right)\left(3,3,0\right)\left(3,3,1\right)\left(3,3,2\right)$)