To find the vector parallel to $\vec{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ with a magnitude of 7, follow these steps:

Step 1: Find the magnitude of the given vector $\vec{u}$.

The magnitude of $\vec{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ is calculated using the formula:

$|\vec{u}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

So, the magnitude of $\vec{u}$ is $\sqrt{14}$.

Step 2: Normalize the vector $\vec{u}$.

To get a unit vector in the direction of $\vec{u}$, divide each component of $\vec{u}$ by its magnitude:

$\hat{u} = \frac{1}{\sqrt{14}} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{14}} \\ \frac{2}{\sqrt{14}} \\ \frac{3}{\sqrt{14}} \end{bmatrix}$

Step 3: Scale the unit vector to the desired magnitude.

Now, multiply the unit vector by 7 to obtain a vector with a magnitude of 7:

$\vec{v} = 7 \times \hat{u} = 7 \times \begin{bmatrix} \frac{1}{\sqrt{14}} \\ \frac{2}{\sqrt{14}} \\ \frac{3}{\sqrt{14}} \end{bmatrix} = \begin{bmatrix} \frac{7}{\sqrt{14}} \\ \frac{14}{\sqrt{14}} \\ \frac{21}{\sqrt{14}} \end{bmatrix}$

Simplifying:

[ \vec{v} = \begin{bmatrix} \frac{7}{\sqrt{14}} \ \frac{14}{\sqrt{14}} \ \frac{21}{\sqrt{14}} \end{bmatrix} = \sqrt{\frac{7}{2}} \cdot expansion making"