The magnitude of a vector is given by the formula (|\vec{A}| = \sqrt{A_x^2 + A_y^2}). Substituting the values of (A_x) and (A_y), we get (|\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5).

Vector addition is performed component-wise. Therefore, (\vec{A} + \vec{B} = (2\hat{i} - 3\hat{j}) + (4\hat{i} + 5\hat{j}) = (2 + 4)\hat{i} + (-3 + 5)\hat{j} = 6\hat{i} + 2\hat{j}).

Vector subtraction is also performed component-wise. Therefore, (\vec{A} - \vec{B} = \langle 2, 3, 4 \rangle\ - \langle -1, 5, -2 \rangle\ = \langle 2 - (-1), 3 - 5, 4 - (-2) \rangle\ = \langle 3, -2, 6 \rangle).

Scalar multiplication is performed by multiplying each component of the vector by the scalar. Therefore, (2\vec{A} = 2(5\hat{i} - 2\hat{j}) = 10\hat{i} - 4\hat{j}).

The dot product of two vectors is given by the formula (\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z). Substituting the values of the components, we get (\vec{A} \cdot \vec{B} = (1)(4) + (2)(-5) + (3)(6) = 4 - 10 + 18 = 12).

The cross product of two vectors is given by the formula (\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix}). Expanding the determinant, we get (\vec{A} \times \vec{B} = \langle (2)(-4) - (-1)(8), (-1)(2) - (1)(4), (1)(-1) - (2)(2) \rangle\ = \langle -11, 2, 5 \rangle).

A vector is perpendicular to two other vectors if its dot product with both of them is zero. Let (\vec{C} = \langle x, y, z \rangle) be the vector we are looking for. Then, (\vec{A} \cdot \vec{C} = (1)(x) + (2)(y) + (-3)(z) = 0) and (\vec{B} \cdot \vec{C} = (2)(x) + (-1)(y) + (4)(z) = 0). Solving these two equations simultaneously, we get (\vec{C} = \langle -1, 1, -1 \rangle).

The angle between two vectors is given by the formula (\theta = \cos^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}\right)). Substituting the values of the vectors, we get (\theta = \cos^{-1}\left(\frac{(1)(1) + (1)(-1)}{\sqrt{1^2 + 1^2}\sqrt{1^2 + (-1)^2}}\right) = \cos^{-1}\left(\frac{0}{\sqrt{2}\sqrt{2}}\right) = \cos^{-1}(0) = 90\degree).

The unit vector in the same direction as (\vec{A}) is given by (\hat{A} = \frac{\vec{A}}{|\vec{A}|}). Substituting the values of (\vec{A}) and (|\vec{A}|), we get (\hat{A} = \frac{3\hat{i} - 4\hat{j}}{\sqrt{3^2 + (-4)^2}} = \frac{3\hat{i} - 4\hat{j}}{5} = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}).

Two vectors are parallel if they have the same direction. This means that one vector can be obtained by multiplying the other vector by a scalar. In this case, we can see that (3\hat{i} - 2\hat{j} = \frac{3}{2}(2\hat{i} - 3\hat{j})). Therefore, (\vec{A}) and (3\hat{i} - 2\hat{j}) are parallel.

The area of the parallelogram formed by two vectors is given by the formula (A = |\vec{A} \times \vec{B}|). Substituting the values of the vectors, we get (A = |\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 0 \ 4 & -1 & 0 \end{vmatrix}| = |\langle 3, 8, -10 \rangle| = \sqrt{3^2 + 8^2 + (-10)^2} = \sqrt{9 + 64 + 100} = \sqrt{173} = 14).

A vector (\vec{C}) is a linear combination of the vectors (\vec{A}) and (\vec{B}) if there exist scalars (a) and (b) such that (\vec{C} = a\vec{A} + b\vec{B}). In this case, we can see that (3\hat{i} - \hat{j} = 3(\hat{i} + 2\hat{j}) + (-1)(2\hat{i} - \hat{j})). Therefore, (\vec{C} = 3\hat{i} - \hat{j}) is a linear combination of (\vec{A}) and (\vec{B}).

The projection of a vector (\vec{A}) onto a vector (\vec{B}) is given by the formula (\text{proj}{\vec{B}}\vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\vec{B}). Substituting the values of the vectors, we get (\text{proj}{\vec{B}}\vec{A} = \frac{(2)(4) + (3)(-1)}{(4^2 + (-1)^2)}(4\hat{i} - \hat{j}) = \frac{8 - 3}{17}(4\hat{i} - \hat{j}) = \frac{5}{17}(4\hat{i} - \hat{j}) = \frac{20}{17}\hat{i} - \frac{5}{17}\hat{j}).

A vector is orthogonal to two other vectors if its dot product with both of them is zero. Let (\vec{C} = \langle x, y, z \rangle) be the vector we are looking for. Then, (\vec{A} \cdot \vec{C} = (1)(x) + (2)(y) + (-3)(z) = 0) and (\vec{B} \cdot \vec{C} = (2)(x) + (-1)(y) + (4)(z) = 0). Solving these two equations simultaneously, we get (\vec{C} = \langle 1, 1, -1 \rangle).