Given vectors a = 2i + j − 3k, b = 4i − j + 2k, and c = i + 3j − k, find the vector AB x AC and the area of triangle ABC.

The position vectors are given as:
$\vec{a} = 2\vec{i} + \vec{j} - 3\vec{k}$
$\vec{b} = 4\vec{i} - \vec{j} + 2\vec{k}$
$\vec{c} = \vec{i} + 3\vec{j} - \vec{k}$

a) Find the vector $\vec{AB} \times \vec{AC}$.

Step 1: Find vector $\vec{AB}$.

$$\vec{AB} = \vec{b} - \vec{a} = (4\vec{i} - \vec{j} + 2\vec{k}) - (2\vec{i} + \vec{j} - 3\vec{k})$$ $$\vec{AB} = (4-2)\vec{i} + (-1-1)\vec{j} + (2-(-3))\vec{k}$$ $$\vec{AB} = 2\vec{i} - 2\vec{j} + 5\vec{k}$$

Step 2: Find vector $\vec{AC}$.

$$\vec{AC} = \vec{c} - \vec{a} = (\vec{i} + 3\vec{j} - \vec{k}) - (2\vec{i} + \vec{j} - 3\vec{k})$$ $$\vec{AC} = (1-2)\vec{i} + (3-1)\vec{j} + (-1-(-3))\vec{k}$$ $$\vec{AC} = -\vec{i} + 2\vec{j} + 2\vec{k}$$

Step 3: Calculate the cross product $\vec{AB} \times \vec{AC}$.

$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -2 & 5 \\ -1 & 2 & 2 \end{vmatrix}$$ $$= \vec{i}((-2)(2) - (5)(2)) - \vec{j}((2)(2) - (5)(-1)) + \vec{k}((2)(2) - (-2)(-1))$$ $$= \vec{i}(-4 - 10) - \vec{j}(4 - (-5)) + \vec{k}(4 - 2)$$ $$= -14\vec{i} - 9\vec{j} + 2\vec{k}$$

The vector $\vec{AB} \times \vec{AC}$ is $\boxed{-14\vec{i} - 9\vec{j} + 2\vec{k}}$.

b) Hence, find the area of the triangle ABC.

Step 1: Calculate the magnitude of $\vec{AB} \times \vec{AC}$.

$$|\vec{AB} \times \vec{AC}| = \sqrt{(-14)^2 + (-9)^2 + (2)^2}$$ $$= \sqrt{196 + 81 + 4}$$ $$= \sqrt{281}$$

Step 2: Calculate the area of triangle ABC.
The area of triangle ABC is half the magnitude of the cross product.

$$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{281}$$

The area of the triangle ABC is $\boxed{\frac{1}{2}\sqrt{281} \text{ units}^2}$.

c) Find the Cartesian equation of the plane ABC.

Step 1: Use the normal vector from part a).
The normal vector to the plane ABC is $\vec{n} = -14\vec{i} - 9\vec{j} + 2\vec{k}$.
The equation of the plane is of the form $n_x x + n_y y + n_z z = d$.
So, $-14x - 9y + 2z = d$.

Step 2: Substitute the coordinates of point A $(2, 1, -3)$ into the equation to find $d$.

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