Solve the following simultaneous equations: 3x+y=7 x−2y=5

$$\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2)$$ $$= 12 - 6 - 2$$ $$= 4$$

$$|\vec{a}| = \sqrt{(2)^2 + (2)^2 + (-1)^2}$$ $$= \sqrt{4 + 4 + 1}$$ $$= \sqrt{9}$$ $$= 3$$

$$|\vec{b}| = \sqrt{(6)^2 + (-3)^2 + (2)^2}$$ $$= \sqrt{36 + 9 + 4}$$ $$= \sqrt{49}$$ $$= 7$$

$$\cos\theta = \frac{4}{(3)(7)}$$ $$\cos\theta = \frac{4}{21}$$ $$\theta = \arccos\left(\frac{4}{21}\right)$$

The angle between the vectors is $\boxed{\arccos\left(\frac{4}{21}\right)}$.

2. Find the projection of the vector $\vec{a} = 2\vec{i} - 3\vec{j} + 6\vec{k}$ on the vector $\vec{b} = \vec{i} + 2\vec{j} + 2\vec{k}$.

The scalar projection of vector $\vec{a}$ onto vector $\vec{b}$ is given by the formula $\text{comp}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

$$\vec{a} \cdot \vec{b} = (2)(1) + (-3)(2) + (6)(2)$$ $$= 2 - 6 + 12$$ $$= 8$$

$$|\vec{b}| = \sqrt{(1)^2 + (2)^2 + (2)^2}$$ $$= \sqrt{1 + 4 + 4}$$ $$= \sqrt{9}$$ $$= 3$$

$$\text{comp}_{\vec{b}}\vec{a} = \frac{8}{3}$$

The projection of $\vec{a}$ on $\vec{b}$ is $\boxed{\frac{8}{3}}$.

3. Find the area of the parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\vec{i} + \vec{j} + 4\vec{k}$ and $\vec{b} = \vec{i} - \vec{j} + \vec{k}$.

The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product, $|\vec{a} \times \vec{b}|$.

$$\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}$$ $$= \vec{i}((1)(1) - (4)(-1)) - \vec{j}((3)(1) - (4)(1)) + \vec{k}((3)(-1) - (1)(1))$$ $$= \vec{i}(1 - (-4)) - \vec{j}(3 - 4) + \vec{k}(-3 - 1)$$ $$= \vec{i}(5) - \vec{j}(-1) + \vec{k}(-4)$$ $$= 5\vec{i} + \vec{j} - 4\vec{k}$$

$$|\vec{a} \times \vec{b}| = \sqrt{(5)^2 + (1)^2 + (-4)^2}$$ $$= \sqrt{25 + 1 + 16}$$ $$= \sqrt{42}$$

The area of the parallelogram is $\boxed{\sqrt{42} \text{ square units}}$.

4. Find the area of the triangle whose adjacent sides are given by the vectors $\vec{a} = 3\vec{i} + \vec{j} + 4\vec{k}$ and $\vec{b} = \vec{i} - \vec{j} + \vec{k}$.

The area of a triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is half the magnitude of their cross product, $\frac{1}{2}|\vec{a} \times \vec{b}|$. We already calculated $\vec{a} \times \vec{b}$ and its magnitude in Question 3.

$$\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}|$$ $$\text{Area} = \frac{1}{2} \sqrt{42}$$

The area of the triangle is $\boxed{\frac{\sqrt{42}}{2} \text{ square units}}$.