Step 1: Understanding the Question:
To find the normal to a plane defined by three points, we can take the cross product of two vectors formed by those points.
Step 2: Key Formula or Approach:
If points are A, B, and C, normal $\vec{n} = \vec{AB} \times \vec{AC}$.
Direction Cosines (DCs) are obtained by normalizing the components of $\vec{n}$.
Step 3: Detailed Explanation:
Let $A = (4, 2, 3), B = (-1, 4, 2), C = (3, 2, 1)$.
Vector $\vec{AB} = (-1-4, 4-2, 2-3) = (-5, 2, -1)$.
Vector $\vec{AC} = (3-4, 2-2, 1-3) = (-1, 0, -2)$.
Normal vector $\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 2 & -1 \\ -1 & 0 & -2 \end{vmatrix}$
$\vec{n} = \hat{i}(-4 - 0) - \hat{j}(10 - 1) + \hat{k}(0 - (-2))$
$\vec{n} = -4\hat{i} - 9\hat{j} + 2\hat{k} = (-4, -9, 2)$.
Magnitude $|\vec{n}| = \sqrt{(-4)^2 + (-9)^2 + 2^2} = \sqrt{16 + 81 + 4} = \sqrt{101}$.
The direction cosines are:
$l = \frac{-4}{\sqrt{101}}, m = \frac{-9}{\sqrt{101}}, n = \frac{2}{\sqrt{101}}$.
Step 4: Final Answer:
The DCs are $(\frac{-4}{\sqrt{101}}, \frac{-9}{\sqrt{101}}, \frac{2}{\sqrt{101}})$.