The problem asks for the value of $\| c\mathbf{u} \|$, where $\mathbf{u} = \langle 5, -12 \rangle$ and $c = -3$.

Solution:

First, $c\mathbf{u}$ represents the scalar multiplication of vector $\mathbf{u}$ by the constant $c$. So,

$c\mathbf{u} = -3 \times \langle 5, -12 \rangle = \langle -15, 36 \rangle$

Now, we need to find the magnitude (norm) of this new vector $\langle -15, 36 \rangle$. The magnitude of a vector $\mathbf{v} = \langle v_1, v_2 \rangle$ is given by:

$\| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2}$

So,

$\| c\mathbf{u} \| = \sqrt{(-15)^2 + 36^2} = \sqrt{225 + 1296} = \sqrt{1521} = 39$

Thus, $\| c\mathbf{u} \| = 39$.

The correct answer is 39.

Do you want more details on this calculation, or have any other questions?

Here are five related questions to further your understanding:

1. What is the general formula for the magnitude of a vector?
2. How does scalar multiplication affect the magnitude of a vector?
3. How would the problem change if $c = 2$?
4. What happens to the direction of a vector when multiplied by a negative scalar?
5. Can the magnitude of a vector ever be negative?

Tip: When multiplying a vector by a scalar, remember that the magnitude scales by the absolute value of the scalar, but the direction reverses if the scalar is negative.