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 . . . . . . .

If the points $P(4, 5, x)$,$Q(3, y, 4)$ and $R(5, 8, 0)$ are collinear,then the value of $x + y$ is

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 $\left| \frac{\vec{AL} + \vec{BM} + \vec{CN}}{\vec{CK}} \right|$.

If $\bar{c} = 5\bar{a} + 6\bar{b}$ and $3\bar{c} = \bar{a} - 4\bar{b}$,then:

$L$ and $M$ are two points with position vectors $2 \vec{a}-\vec{b}$ and $\vec{a}+2 \vec{b}$ respectively. The position vector of the point $N$ which divides the line segment $LM$ in the ratio $2:1$ externally is

If $\overline{a}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}$,$\overline{b}=4 \hat{\imath}+5 \hat{\jmath}+3 \hat{k}$ and $\overline{c}=6 \hat{\imath}+\hat{\jmath}+5 \hat{k}$ are the position vectors of the vertices of a triangle $ABC$ respectively,then the position vector of the intersection of the medians (centroid) of the triangle $ABC$ is: