The area of the quadrilateral $ABCD$ with vertices $A(0,4,1)$,$B(2,3,-1)$,$C(4,5,0)$,and $D(2,6,2)$ is equal to

If a non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ and the plane determined by the vectors $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$,then the angle between the vectors $\vec{a}$ and $\hat{i}+\hat{j}+\hat{k}$ is

The area of the parallelogram whose adjacent sides are $\vec{a} = \hat{i} - \hat{k}$ and $\vec{b} = 2\hat{j} + 3\hat{k}$ is

Let $a$,$b$,and $c$ be real numbers such that $a^2 + b^2 + c^2 = 1$. Then the maximum value of $(4b - 3c)^2 + (4a - 2c)^2 + (3a - 2b)^2$ is:

The area of a parallelogram whose adjacent sides are given by the vectors $i + 2j + 3k$ and $-3i - 2j + k$ (in square units) is

$A$ vector which is orthogonal to the vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and coplanar with the vectors $\vec{b} = 3\hat{i} + 2\hat{j}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$ is