Given the vectors $\vec x = 3i - 6j - k$,$\vec y = i + 4j - 3k$,and $\vec z = 3i - 4j - 12k$,find the projection of the vector $\vec x \times \vec y$ onto the vector $\vec z$.

The parallelepiped is determined by vectors $\vec{a} = -2\hat{i} + 5\hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\vec{c} = -3\hat{i} + \hat{j} + 4\hat{k}$. The altitude of the parallelepiped on the parallelogram base determined by vectors $\vec{b}$ and $\vec{c}$ is

Let $\vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{CB} = -3\hat{i} - 2\hat{k}$ be the diagonals of the parallelogram $OBDC$. If $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$, then the volume of the parallelepiped determined by vectors $\vec{OA}, \vec{OB}$, and $\vec{OC}$ (in cubic units) is:

The value of $x$ for which the volume of the parallelepiped formed by the vectors $\vec{a} = \hat{i} + x\hat{j} + \hat{k}$, $\vec{b} = \hat{j} + x\hat{k}$, and $\vec{c} = x\hat{i} + \hat{k}$ is minimum, is

If $a \cdot b = b \cdot c = c \cdot a = 0$,then what is the value of the scalar triple product $[a b c]$?

Let $x_{0}$ be the point of local maxima of $f(x)=\vec{a} \cdot(\vec{b} \times \vec{c}),$ where $\vec{a}=x \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{b}=-2 \hat{i}+x \hat{j}-\hat{k}$ and $\vec{c}=7 \hat{i}-2 \hat{j}+x \hat{k}$. Then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ at $x=x_{0}$ is: