The problem asks us to find all vectors that are perpendicular to both vector $v_1 = (1, -2, -3)$ and vector $v_2 = (-3, 2, 0)$.

1. Problem Description

The problem asks us to find all vectors that are perpendicular to both vector $v_1 = (1, -2, -3)$ and vector $v_2 = (-3, 2, 0)$.

2. Solution Steps

A vector perpendicular to two given vectors is parallel to their cross product.

Let $v_1 = (1, -2, -3)$ and $v_2 = (-3, 2, 0)$.

The cross product of $v_1$ and $v_2$ is given by:

$$v_1 \times v_2 = \begin{vmatrix} i & j & k \\ 1 & -2 & -3 \\ -3 & 2 & 0 \end{vmatrix} = i((-2)(0) - (-3)(2)) - j((1)(0) - (-3)(-3)) + k((1)(2) - (-2)(-3))$$

$$v_1 \times v_2 = i(0 + 6) - j(0 - 9) + k(2 - 6) = 6i + 9j - 4k$$

So, $v_1 \times v_2 = (6, 9, -4)$.

Any vector perpendicular to both $v_1$ and $v_2$ must be a scalar multiple of $(6, 9, -4)$. Thus, all such vectors have the form $c(6, 9, -4)$, where $c$ is any scalar.

3. Final Answer

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