For $'S_1'$ plane
1. $|x - 1| \le 2$
A. when $(x-1) \ge0 \Rightarrow x \ge 1$
Then $+(x - 1) \le 2 \Rightarrow x \le 3$ $\{$ here $(x \ge 1) \cap (x \le 3) \Rightarrow 1 \le x \le 3 \}$
B. When $(x -1) < 0 \Rightarrow x < 1$
Then $-(x - 1) \le 2 \Rightarrow x \ge -1$ $\{$ here $(x <1) \cap (x \ge -1) \Rightarrow -1 \le x < 1 \}$
Union $(A \cup B) \Rightarrow -1 \le x \le 3$
2. $|y + 2| \le 3$
A. when $(y+2) \ge0 \Rightarrow y \ge -2$
Then $(y + 2) \le 3 \Rightarrow y \le 1$ $\{$ here $(y \ge -2) \cap (y \le 1) \Rightarrow -2 \le y \le 1 \}$
B. When $(y+2) < 0 \Rightarrow y < -2$
Then $-(y+2) \le 3 \Rightarrow y \ge -5$ $\{$ here $(y <-2) \cap (y\ge -5) \Rightarrow -5 \le y < -2 \}$
Union $(A \cup B) \Rightarrow -5 \le y \le 1$
So, $ 'S_1'$ plane will be for intervals : $ -1 \le x \leq 3 \text{ and } -5 \le y \leq 1$
For $'S_2' $plane ,
1. $x - y \ge -2 \Rightarrow y \le x +2$
2. $y \ge 1$
3. $x \le 3$
Here, $S = S_{1} \cup S_2$
So, Area of $S=$ Area of $S_1$ $+$ Area of $S_2$
$\Rightarrow$ Area of $S = 6*4 + \frac{1}{2}*4*4$
$\Rightarrow$ Area of $S= 24 + 8$
$\Rightarrow$ Area of $S= 32$
So, Answer is (C)