In $\triangle ABC$,$L, M, N$ are points on $BC, CA, AB$ respectively,dividing them in the ratio $1:2, 2:3, 3:5$. If the point $K$ divides $AB$ in the ratio $5:3$,then find the value of $\left| \frac{\vec{AL} + \vec{BM} + \vec{CN}}{\vec{CK}} \right|$.

In $\Delta ABC$,if $A(2, 4, -1)$,$B(4, 5, 1)$,and $C(3, 6, -3)$,then $\Delta ABC$ is a ... triangle.

If $ABCDEF$ is a regular hexagon,then $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = $

Given vectors $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$. The vector in the direction of $\vec{a} + \vec{b}$ with magnitude $\sqrt{2}$ is . . . . . . .

Let $A, B, C$ be three points whose position vectors are $\overrightarrow{a} = \hat{i} + 4\hat{j} + 3\hat{k}$,$\overrightarrow{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$ (where $\alpha \in R$),and $\overrightarrow{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear,then the length of the median in $\triangle ABC$ through $A$ is:

If the position vectors of the points $A, B, C, D$ are $2i + 3j + 5k, i + 2j + 3k, -5i + 4j - 2k$ and $i + 10j + 10k$ respectively,then: