$F=(X+Y+Z)(\overline X +Y)(\overline Y +Z)$
Taking complement of above expression;
$\overline {F}=\overline {(X+Y+Z)(\overline X+Y)(\overline Y+Z)}$
Applying De-Morgan’s law;
$\overline {F}=\overline{(X+Y+Z)}+\overline{(\overline X+Y)}+\overline{(\overline Y+Z)}$
$\overline{F}=(\bar X.\bar Y.\bar Z)+\overline{\overline{X}}.\overline Y+\overline{\overline{Y}}.\overline Z$
$\left [\because \overline{\overline{X}}=X, \text{Using double negation law} \right ]$
$\therefore \overline F=(\bar X.\bar Y.\bar Z)+(X.\overline Y)+(Y.\overline Z)$ $\quad \quad \to \text{Option (D)}$
Taking $\overline Y$ as common we get;
$\overline F=\overline Y \left[ (\bar X \bar Z)+X\right]+Y\overline Z$
$\left [ \because A+BC=(A+B)(A+C) \text{ Applying distributive law here}\right ]$
$\overline F=\overline Y\left[(X+\overline X)(X+\overline Z)\right]+Y\overline Z$
$\overline F=\overline Y\left[X+\overline Z\right]+Y\overline Z$
$\overline F=X \overline Y+\overline Y\overline Z+Y\overline Z$
Taking $\overline Z$ as common
$\overline F=X\overline Y+\overline Z(Y+ \overline Y)$
$\because \text{(Y+$\overline Y$=1) using complement law}$
$\therefore \overline F=X\overline Y+\overline Z$$\quad \quad \to \text{Option (B)}$
Applying distributive law here we get;
$\overline F=(X+\overline Z)(\overline Y+\overline Z)$$\quad \quad \to \text{Option (C)}$
So, correct options are $B,C,D.$
Option A is false and can be proved as follows:
Take $X = 0, Y = 1, Z = 0$
Now, $F = 0,$ since $(\bar Y + Z)$ term will be zero. So, $\bar F$ must be $1.$
But option A gives $0$ as the term $X + \bar Y$ evaluates to $0.$ So, option A is not equal to $\bar F.$
Properties of Boolean Algebra