If one root of $24x^3-14x^2-63x+45=0$  is double the other, then the roots are

$\alpha ,2\alpha ,\beta$ are the roots

 sum: $3\alpha +\beta =\frac{7}{12}\to \left(1\right)$  

product of two roots: $2{\alpha }^2+3\alpha \beta =\frac{-21}{8}\to \left(2\right)$  

From equations 1 and 2

 $\Rightarrow 8{\alpha }^2-2\alpha -3=0$

$\Rightarrow \alpha =\frac{3}{4},2\alpha =\frac{3}{2},\beta =\frac{-5}{3}$           or (By verification answer : C)