Find the area of the triangle with vertices $A(1,1,2), B(2,3,5)$ and $C(1,5,5)$.

Let $L_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}$ and $L_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $L_1$ and $L_2$ is

If $|\overrightarrow{a}|=2, |\overrightarrow{b}|=3$ and $\overrightarrow{a}, \overrightarrow{b}$ are mutually perpendicular,then the area of the triangle whose vertices are $\overrightarrow{0}, \overrightarrow{a}+\overrightarrow{b}, \overrightarrow{a}-\overrightarrow{b}$ is

If $a \times b = b \times c \ne 0,$ where $a, b$ and $c$ are coplanar vectors,then for some scalar $k$

If $|\bar{u}| = 8$ and $|\bar{v}| = 12$ with an angle of $150^{\circ}$ between them,then find $|\bar{u} \times \bar{v}|$.

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: