The straight line $\frac{x+3}{3} = \frac{y+2}{4} = \frac{z+1}{0}$ is

The line is given as $\frac{x+3}{3} = \frac{y+2}{4} = \frac{z+1}{0}$

This is the symmetric form of a line in 3D space. The standard format is:

$\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$

where $(a, b, c)$ is the direction vector of the line.

---

From the equation $\frac{x+3}{3} = \frac{y+2}{4} = \frac{z+1}{0}$, the numbers in the denominators give the direction vector:

Direction vector $= (3, 4, 0)$

The z-component is 0, which means the line does not move in the z-direction at all. The line stays at a constant z-value of $z = -1$ and is completely flat with respect to the z-axis.

---

Direction vectors of the axes:

x-axis: $(1, 0, 0)$

y-axis: $(0, 1, 0)$

z-axis: $(0, 0, 1)$

Line's direction vector: $(3, 4, 0)$

---

For the line to be parallel to an axis, the direction vectors must be proportional.

$(3, 4, 0)$ is not proportional to $(1, 0, 0)$, so not parallel to x-axis.

$(3, 4, 0)$ is not proportional to $(0, 1, 0)$, so not parallel to y-axis.

$(3, 4, 0)$ is not proportional to $(0, 0, 1)$, so not parallel to z-axis.

---

For two lines to be perpendicular, their dot product must equal 0.

Dot product $= (3, 4, 0) \cdot (0, 0, 1)$

$= (3 \times 0) + (4 \times 0) + (0 \times 1)$

$= 0$

The line is perpendicular to the z-axis.

---

When the z-component of the direction vector equals 0, the line is automatically perpendicular to the z-axis, since the line doesn't move in that direction at all.

Therefore, the line is perpendicular to the z-axis.