Planck’s constant h, speed of light  c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option is (are)

$\begin{array}{l}Mass in terms of h , c , G\\ \mathrm{M}\propto {\mathrm{h}}^{\mathrm{a}}{\mathrm{c}}^b{\mathrm{G}}^c\\ \Rightarrow {\mathrm{M}}^1\propto {\left({\mathrm{ML}}^2{\mathrm{T}}^{-1}\right)}^{\mathrm{a}}{\left({\mathrm{LT}}^{-1}\right)}^{\mathrm{b}}{\left({\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right)}^{\mathrm{c}}\\ \Rightarrow {\mathrm{M}}^1\propto {\mathrm{M}}^{\mathrm{a}-c}{\mathrm{L}}^{2\mathrm{a}+\mathrm{b}+3\mathrm{c}}{\mathrm{T}}^{-\mathrm{a}-\mathrm{b}-2\mathrm{c}}\\ Comparing powers,\\ \mathrm{a}-\mathrm{c}=1\to \left(1\right)\\ 2\mathrm{a}+\mathrm{b}+3\mathrm{c}=0\to \left(2\right)\\ \mathrm{a}+\mathrm{b}+2\mathrm{c}=0\to \left(3\right)\end{array}$

On solving (1), (2), (3),  $a=\frac{1}{2},b=\frac{1}{2},c=-\frac{1}{2}$
$\therefore M\propto \sqrt{c}$  only $\to \left(A\right)$  is correct
In the same way

  $\begin{array}{l}Length in terms of h , c , G\\ \text{L}\propto {\text{h}}^{\text{a}}{\text{c}}^b{\text{G}}^c\\ \Rightarrow L^1\propto {\left(M{L}^2{\text{T}}^{-1}\right)}^{\text{a}}{\left(L{T}^{-1}\right)}^{\text{b}}{\left({\text{M}}^{-1}{\text{L}}^3{\text{T}}^{-2}\right)}^{\text{c}}\\ \Rightarrow L^1\propto {\text{M}}^{\text{a}-c}{\text{L}}^{2\text{a}+\text{b}+3\text{c}}{\text{T}}^{-\text{a}-\text{b}-2\text{c}}\\ \begin{array}{l}Comparing powers,\\ \text{a}-\text{c}=0\to \left(1\right)\\ 2\text{a}+\text{b}+3\text{c}=1\to \left(2\right)\\ \text{a}+\text{b}+2\text{c}=0\to \left(3\right)\end{array}\\ \begin{array}{l}After Solving \\ a=\frac{1}{2} , b=\frac{-3}{2} , c=\frac{1}{2}\end{array}\\ \Rightarrow L\propto h^{\frac{1}{2}}{c}^{\frac{-3}{2}}{G}^{\frac{1}{2}}\end{array}$
$L\propto \sqrt{h},L\propto \sqrt{G}\to \left(C\right),\left(D\right)$  are correct