$\text{ D }\mathrm{ABC}\text{ is a tetrahedron with }\mathrm{A}=(2,0,0),\mathrm{B}=(0,4,0)\text{ and }\mathrm{CD}=\sqrt{14}\text{. Edge }\mathrm{CD}\text{ lies }$$\text{ on the line }\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\text{. If locus of centroid of tetrahedron is }$$\frac{x-\frac{3}{2}}{1}=\frac{y-y_1}{a}=\frac{z-z_1}{b},\text{ then which of the following is/are true: }$

$\text{ Let }\mathrm{C}=\left({\mathrm{\lambda }}_1+1,2{\mathrm{\lambda }}_1+2,3{\mathrm{\lambda }}_1+3\right)\text{ and }\mathrm{D}=\left({\mathrm{\lambda }}_2+1,2{\mathrm{\lambda }}_2+2,3{\mathrm{\lambda }}_2+3\right)$

$\begin{array}{l}\therefore \mathrm{CD}=\sqrt{14}\left({\mathrm{\lambda }}_1-{\mathrm{\lambda }}_2\right)=\sqrt{14}\Rightarrow {\mathrm{\lambda }}_1={\mathrm{\lambda }}_2+1\\ \Rightarrow \text{ Let centroid }(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })=\left(\frac{5+2{\mathrm{\lambda }}_2}{4},\frac{10+4{\mathrm{\lambda }}_2}{4},\frac{9+6{\mathrm{\lambda }}_2}{4}\right)\\ \therefore \text{ Locus is }\frac{4\mathrm{x}-5}{2}=\frac{4\mathrm{y}-10}{4}=\frac{4\mathrm{z}-9}{6}\end{array}$

$\therefore \text{ Dr's of line }1,2,3\text{ and point }\left(\frac{3}{2},3,3\right)\text{ satisfies the line }$

$\therefore \text{ locus is }\frac{\mathrm{x}-\frac{3}{2}}{1}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-3}{3}$