We need to find the unit vectors perpendicular to the plane determined by the points $(1, 3, 5)$, $(3, -1, 2)$, and $(4, 0, 1)$.

1. Problem Description

We need to find the unit vectors perpendicular to the plane determined by the points $(1, 3, 5)$, $(3, -1, 2)$, and $(4, 0, 1)$.

2. Solution Steps

First, let's find two vectors in the plane. We can do this by subtracting the coordinates of the given 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 $\vec{AC} = C - A = (4-1, 0-3, 1-5) = (3, -3, -4)$.

Now, we find a vector normal to the plane by taking the cross product of $\vec{AB}$ and $\vec{AC}$.

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

The normal vector is $\vec{n} = (7, -1, 6)$.

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

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

To find the unit vectors, we divide the normal vector by its magnitude:

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

3. Final Answer

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