Analyze the first clue: red solid is insoluble in water, but dissolves when KI is added

A compound that is insoluble in water but dissolves in excess iodide ion often does so by forming a soluble complex.

For mercuric iodide: $HgI_2(s) + 2I^- \rightarrow [HgI_4]^{2-}$

Since $KI$ provides $I^-$ ions, red $HgI_2$ dissolves in excess $KI$ due to formation of soluble tetraiodomercurate(II) complex.

Analyze the second clue: on heating, violet fumes are liberated and metal droplets appear

Violet fumes strongly suggest iodine vapors, $I_2$.

If $HgI_2$ is heated strongly, it decomposes: $HgI_2 \rightarrow Hg + I_2$

Here:

• $I_2$ gives violet fumes
• $Hg$ condenses as metal droplets on cooler parts of the test tube

This matches the observation exactly.

Check the other options

B: $HgO$

• On heating: $2HgO \rightarrow 2Hg + O_2$
• This gives mercury droplets, but no violet fumes.
• Also, dissolution in $KI$ is not the characteristic clue here.
• So, not correct.

C: $Pb_3O_4$

• Red lead is red, but on heating it does not produce violet iodine fumes or mercury droplets.
• It also does not dissolve in water upon adding $KI$ in the described way.
• Not correct.

D: $(NH_4)_2Cr_2O_7$

• This orange-red solid decomposes on heating as: $(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + N_2 + 4H_2O$
• No violet fumes, no metal droplets.
• Not correct.

Conclusion

The red solid is: $\boxed{HgI_2}$

So the correct option is A.