The area of the triangle with vertices $(1,2,0)$,$(1,0,2)$ and $(0,3,1)$ is.

If $a = i - j$,$b = i + j$,$c = i + 3j + 5k$ and $n$ is a unit vector such that $b \cdot n = 0$ and $a \cdot n = 0$,then the value of $|c \cdot n|$ is equal to

Let $\vec{a}=6 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $|\vec{c}| \geq 6$,$\vec{a} \cdot \vec{c}=6|\vec{c}|$,$|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^{\circ}$,then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

Given $a = i + j - k$,$b = -i + 2j + k$,and $c = -i + 2j - k$. $A$ unit vector perpendicular to both $a + b$ and $b + c$ is

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 $a=$

The unit vector perpendicular to both $i + j$ and $j + k$ is