If velocity, force and time are taken as the fundamental...

Let the quantity be Q, then,

$Q = f ( v, F, T )$

Assuming that the function is the product of power functions of $v, F$ and $T$,

$Q = Kv^{x}F^{y}T^{z}... ( i )$ & &

where K is a dimensionless constant of proportionality. The above equation dimensionally becomes

$[ Q ] = [ LT^{- 1}{]}^{x} [ MLT^{- 2}{]}^{y} [ T{]}^{z}$

i.e., $[ Q ] = [ M^{y}L^{( x + y )}T^{( - x - 2y + z )} ]$, Now

Q = mass i.e., $[ Q ] = [ M ]$

So Equation (ii) becomes

$[ M ] = [ M^{y}L^{( x + y )}T^{( - x - 2y + z )} ]$
its dimensional correctness requires

$y = 1, x + y = 0$and $- x - 2y + z = 0$

which on solving yields

$x = - 1, y = 1$and $z = 1$

Substituting it in Equation (i), we get

$Q = Kv^{- 1}FT$