Let A = \ x ∈ R : |x - a| 0 and d > 0 . If A B = (1, 2] [6, 7) ,...

We are given $A = \{x \in R : |x - a| < b\}$ and $b > 0$.
$\Rightarrow -b < x - a < b$
$\Rightarrow x \in (a - b, a + b)$

We are also given $B = \{x \in R : |x - c| \ge d\}$ and $d > 0$.
$\Rightarrow x - c \le -d$ or $x - c \ge d$
$\Rightarrow x \in (-\infty, c - d] \cup [c + d, \infty)$

The intersection of sets $A$ and $B$ is:
$A \cap B = (a - b, c - d] \cup [c + d, a + b)$

We are given that $A \cap B = (1, 2] \cup [6, 7)$.
Comparing the intervals, we get:
$a - b = 1$
$c - d = 2$
$c + d = 6$
$a + b = 7$

Solving these equations:
Adding the first and fourth equations: $2a = 8 \Rightarrow a = 4$.
Substituting $a = 4$ in the first equation: $4 - b = 1 \Rightarrow b = 3$.
Adding the second and third equations: $2c = 8 \Rightarrow c = 4$.
Substituting $c = 4$ in the second equation: $4 - d = 2 \Rightarrow d = 2$.

Finally, the value of $a + 2b + 3c + 4d$ is:
$4 + 2(3) + 3(4) + 4(2) = 4 + 6 + 12 + 8 = 30$.

Answer: $30$