Find a unit vector perpendicular to both vectors $3i + 2j - k$ and $12i + 5j - 5k$.

Using vectors,prove that parallelograms on the same base and between the same parallels are equal in area.

The area of the parallelogram with vertices $A(1, 2, 3)$,$B(1, 3, a)$,$C(3, 8, 6)$,and $D(3, 7, 3)$ is $\sqrt{265}$ sq. units. Then $a=$

If the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is $15$ sq. units,then the area of the parallelogram having $3 \vec{a}+\vec{b}$ and $\vec{a}+3 \vec{b}$ as two adjacent sides,in square units,is

$A$ vector of length $3$ perpendicular to each of the vectors $3\,i + j - 4\,k$ and $6\,i + 5\,j - 2\,k$ is

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$,where $|\vec{a}|=4, |\vec{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^{2} + 4(\vec{a} \cdot \vec{b})^{2}$ is equal to