$\mathrm{P}$ (exactly one of $A$ or $B$ occurs)

$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$      ….$(1)$

$\mathrm{P}(\text { Exactly one of } \mathrm{B} \text { or $C$ occurs })$

$=P(B)+P(C)-2 P(B \cap C)=\frac{1}{4}$        …..$(2)$

P (Exactly one of $C$ or $A$ occurs)

$=\mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A})-2 \mathrm{P}(\mathrm{C} \cap \mathrm{A})=\frac{1}{4}$          …..$(3)$

Adding $(1),(2)$ and $(3),$ we get

$2 \Sigma P(A)-2 \Sigma P(A \cap B)=\frac{3}{4}$

$\therefore \Sigma P(A)-\Sigma P(A \cap B)=\frac{3}{8}$

Now, $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=\frac{1}{16}$

$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})$

$=\Sigma P(A)-\Sigma P(A \cap B)+P(A \cap B \cap C)$

$=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$