A body performs uniform circular motion with radius $\vec{r} = 4\hat{i} - 2\hat{j} + 2\hat{k}$ and angular velocity $\vec{\omega} = 2\hat{i} - 3\hat{j} + \hat{k}$. The magnitude of linear velocity of the body is

The linear velocity $\vec{v}$ of a body performing uniform circular motion is related to its angular velocity $\vec{\omega}$ and radius vector $\vec{r}$ by the cross product relation:

$\vec{v} = \vec{\omega} \times \vec{r}$

Given the angular velocity and radius vectors:
$\vec{\omega} = 2\hat{i} - 3\hat{j} + \hat{k}$
$\vec{r} = 4\hat{i} - 2\hat{j} + 2\hat{k}$

The cross product is evaluated using the determinant method:

$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 4 & -2 & 2 \end{vmatrix}$

Expanding the determinant along the first row:

$\vec{v} = \hat{i} [(-3)(2) - (1)(-2)] - \hat{j} [(2)(2) - (1)(4)] + \hat{k} [(2)(-2) - (-3)(4)]$

$\vec{v} = \hat{i} [-6 + 2] - \hat{j} [4 - 4] + \hat{k} [-4 + 12]$

$\vec{v} = -4\hat{i} + 8\hat{k}$

The magnitude of the linear velocity is given by:

$|\vec{v}| = \sqrt{(-4)^2 + (8)^2}$

$|\vec{v}| = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$

Answer: $4\sqrt{5}$