Planck' constant h , speed of light c and gravitational constant G...

$\left[h\right] =$\left[M{L}^{2}{T}^{- 1}\right]$<br >$\left[c\right] = $\left[L{T}^{- 1}\right]$
$\left[G\right] =$\left[{M}^{- 1}{L}^{3}{T}^{- 2}\right]$<br >For unit of length<br >Let$L ∝ {h}^{x}{c}^{y}{G}^{z}$<br >$⇒$From principle of homogeneity<br >$\left[LHS\right] = $\left[RHS\right]$
$⇒$\left[{M}^{0}L{T}^{0}\right] = {\left[M{L}^{2}{T}^{- 1}\right]}^{x}{\left[L{T}^{- 1}\right]}^{y}{\left[{M}^{- 1}{L}^{3}{T}^{- 2}\right]}^{z}
${M}^{0}L{T}^{0} = {M}^{x - z}{L}^{2x + y + 3z}{T}^{- x - y - 2z}$
Comparing we get
$x - z = 0$
$2x + y + 3z = 1$
$- x - y - 2z = 0$
Solving we get, $x = \dfrac{1}{2}, y = \dfrac{- 3}{2}, z = \dfrac{1}{2}$
$L = \sqrt{\dfrac{hG}{{c}^{3}}}$
For unit of mass
Let $M ∝ {h}^{x}{c}^{y}{G}^{z}$
Solving in similar manner as above
We get $x = \dfrac{1}{2}, y = \dfrac{1}{2}, z = \dfrac{- 1}{2}$
$M = \sqrt{\dfrac{hc}{G}}$