The problem asks us to find the unit vectors perpendicular to the plane determined by the three points (1, 3, 5), (3, -1, 2), and (4, 0, 1).

1. Problem Description

The problem asks us to find the unit vectors perpendicular to the plane determined by the three points (1, 3, 5), (3, -1, 2), and (4, 0, 1).

2. Solution Steps

First, we need to find two vectors lying in the plane. We can do this by subtracting the coordinates of the points. Let $A = (1, 3, 5)$, $B = (3, -1, 2)$, and $C = (4, 0, 1)$.

Vector $\vec{AB} = B - A = (3-1, -1-3, 2-5) = (2, -4, -3)$.

Vector $\vec{AC} = C - A = (4-1, 0-3, 1-5) = (3, -3, -4)$.

Next, we find a vector normal to the plane by taking the cross product of these two vectors.

The cross product $\vec{n} = \vec{AB} \times \vec{AC}$ is given by:

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & -3 \\ 3 & -3 & -4 \end{vmatrix} = \hat{i}((-4)(-4) - (-3)(-3)) - \hat{j}((2)(-4) - (-3)(3)) + \hat{k}((2)(-3) - (-4)(3))$

$\vec{n} = \hat{i}(16 - 9) - \hat{j}(-8 + 9) + \hat{k}(-6 + 12) = 7\hat{i} - \hat{j} + 6\hat{k} = (7, -1, 6)$

Now, we need to find the magnitude of this normal vector:

$||\vec{n}|| = \sqrt{7^2 + (-1)^2 + 6^2} = \sqrt{49 + 1 + 36} = \sqrt{86}$

Finally, we find the unit vectors by dividing the normal vector by its magnitude. There are two such vectors, one in each direction:

$\hat{u} = \pm \frac{\vec{n}}{||\vec{n}||} = \pm \frac{(7, -1, 6)}{\sqrt{86}} = \pm (\frac{7}{\sqrt{86}}, \frac{-1}{\sqrt{86}}, \frac{6}{\sqrt{86}})$

3. Final Answer

The unit vectors perpendicular to the plane are $(\frac{7}{\sqrt{86}}, \frac{-1}{\sqrt{86}}, \frac{6}{\sqrt{86}})$ and $(-\frac{7}{\sqrt{86}}, \frac{1}{\sqrt{86}}, -\frac{6}{\sqrt{86}})$.