Given $$P\left( A \right) = {{3x + 1} \over 3},$$ $$P\left( B \right) = {{1 - x} \over 4}$$ and $$P\left( C \right) = {{1 - 2x} \over 2}$$

We know for any event X, $$0 \le P\left( X \right) \le 1$$

$$\therefore$$ $$0 \le {{3x + 1} \over 3} \le 1$$
$$ \Rightarrow - 1 \le 3x \le 2$$
$$ \Rightarrow - {1 \over 3} \le x \le {2 \over 3}$$

$$0 \le {{1 - x} \over 4} \le 1$$
$$ \Rightarrow - 3 \le x \le 1$$

$$0 \le {{1 - 2x} \over 2} \le 1$$
$$ \Rightarrow - 1 \le 2x \le 1$$
$$ \Rightarrow - {1 \over 2} \le x \le {1 \over 2}$$

In the question given that A, B and C are mutually exclusive
So $$P\left( {A \cup B \cup C} \right)$$ = $$P\left( {A} \right)$$ + $$P\left( {B} \right)$$ + $$P\left( {C} \right)$$
 
$$ \Rightarrow P\left( {A \cup B \cup C} \right)$$ = $${{3x + 1} \over 3}$$ + $${{1 - x} \over 4}$$ + $${{1 - 2x} \over 2}$$

$$\therefore$$ 0 $$ \le $$ $${{3x + 1} \over 3}$$ + $${{1 - x} \over 4}$$ + $${{1 - 2x} \over 2}$$ $$ \le $$ 1

0 $$ \le $$ 13 - 3$$x$$ $$ \le $$ 12

$$ \Rightarrow {1 \over 3} \le x \le {{13} \over 3}$$

From all those relations, we get

$$\max \left\{ { - {1 \over 3}, - 3, - {1 \over 2},{1 \over 3}} \right\}$$ $$ \le $$ $$x$$ $$ \le $$ $$\min \left\{ {{2 \over 3},1,{1 \over 2},{{13} \over 3}} \right\}$$

So, $${1 \over 3} \le x \le {1 \over 2}$$

$$ \Rightarrow $$ $$ \Rightarrow x \in \left[ {{1 \over 3},{1 \over 2}} \right]$$