In a $\triangle ABC$,$|CB|=a$,$|CA|=b$,$|AB|=c$ and $CD$ is the median through the vertex $C$. Then,$CA \cdot CD=$

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \cdot(\overline{b}+\overline{c})+\overline{b} \cdot(\overline{c}+\overline{a})+\overline{c} \cdot(\overline{a}+\overline{b})=0$ and $|\overline{a}|=1$,$|\overline{b}|=8$ and $|\overline{c}|=4$,then $|\overline{a}+\overline{b}+\overline{c}|$ has the value

If $\vec{a}$ and $\vec{b}$ are unit vectors,then the maximum value of $|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$ is:

If the angles between the sides of the triangle $ABC$ formed by $A(2,3,5)$,$B(-1,3,2)$ and $C(3,5,-2)$ are $\alpha, \beta$ and $\gamma$,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = $

If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0 .$ But the converse need not be true. Justify your answer with an example.

Vectors $a, b, c$ are inclined to each other at an angle of $60^\circ$. If $|a| = 2, |b| = 2$,and $|c| = 2$,then calculate the value of $(2a + 3b - 5c) \cdot (4a - 6b + 10c)$.