The number of lines passing through A (3, 4) and the sum of whose non-zero intercepts is zero is (are)

Let the equation of the line be $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ and $b$ are the non-zero intercepts on the x-axis and y-axis respectively.

Given that the sum of the non-zero intercepts is zero, we have:
$a + b = 0 \Rightarrow b = -a$

Substituting $b = -a$ into the equation of the line, we get:
$\dfrac{x}{a} + \dfrac{y}{-a} = 1$
$x - y = a$

Since the line passes through the point $A(3, 4)$, we substitute $x = 3$ and $y = 4$ into the equation:
$3 - 4 = a \Rightarrow a = -1$

Since $a = -1 \neq 0$, this is a valid non-zero intercept. The corresponding value of $b$ is $b = -(-1) = 1$.

Thus, there is exactly one such line, and its equation is $x - y = -1$ (or $x - y + 1 = 0$).

Therefore, the number of such lines is $1$.

Answer: $1$