If $ABCD$ is a parallelogram and the position vectors of $A, B, C$ are $i + 3j + 5k, i + j + k$ and $7i + 7j + 7k$, then the position vector of $D$ will be:

Let $\alpha, \beta, \gamma$ be distinct real numbers. The points with position vectors $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$,and $\gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$ form:

For a triangle $ABC$,if $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$,then the length of the median drawn from vertex $A$ is:

If a parallelogram is constructed on the vectors $\vec{a} = 3\vec{p} - \vec{q}$ and $\vec{b} = \vec{p} + 3\vec{q}$, where $|\vec{p}| = 3$, $|\vec{q}| = 2$ and the angle between $\vec{p}$ and $\vec{q}$ is $\pi/3$, then the ratio of the lengths of adjacent sides $|\vec{a}|$ and $|\vec{b}|$ of the parallelogram is:

If $P \hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, and $4 \hat{i}+13 \hat{j}-18 \hat{k}$ are the position vectors of three collinear points $A$, $B$, and $C$ respectively, then the vector in the direction of $AB$ of length $|P|$ units is

The direction cosines of the resultant of the vectors $(i + j + k)$,$(-i + j + k)$,$(i - j + k)$ and $(i + j - k)$ are