The problem asks us to find all vectors that are perpendicular to both $v = (1, -2, -3)$ and $w = (-3, 2, 0)$.

1. Problem Description

The problem asks us to find all vectors that are perpendicular to both $v = (1, -2, -3)$ and $w = (-3, 2, 0)$.

2. Solution Steps

A vector $u = (x, y, z)$ is perpendicular to both $v$ and $w$ if and only if the dot product of $u$ with each of $v$ and $w$ is zero. This gives us two equations:

$u \cdot v = 0$

$u \cdot w = 0$

Writing these out explicitly, we have:

$x - 2y - 3z = 0$

$-3x + 2y = 0$

From the second equation, we get $2y = 3x$, or $y = \frac{3}{2}x$.

Substituting this into the first equation, we have:

$x - 2(\frac{3}{2}x) - 3z = 0$

$x - 3x - 3z = 0$

$-2x - 3z = 0$

$3z = -2x$

$z = -\frac{2}{3}x$

Thus, $u = (x, \frac{3}{2}x, -\frac{2}{3}x)$.

We can factor out an $x$ to get $u = x(1, \frac{3}{2}, -\frac{2}{3})$.

To eliminate fractions, we can multiply the vector by 6 to get $x(6, 9, -4)$. Since $x$ can be any scalar, we can simply say that the set of all vectors perpendicular to $v$ and $w$ are scalar multiples of the vector $(6, 9, -4)$.

3. Final Answer

The vectors perpendicular to both $(1, -2, -3)$ and $(-3, 2, 0)$ are of the form $t(6, 9, -4)$, where $t$ is any real number.