If $A, B, C$ and $D$ are four points in the plane such that $|AB|^2+|CD|^2=|BC|^2+|DA|^2=100$,then $AC \cdot BD=$

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$ be three vectors. If $\vec{v}$ is a vector in the plane of $\vec{a}$ and $\vec{b}$ such that the projection of $\vec{v}$ on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,then $\vec{v} = $

Show that the points $A (2 \hat{i}-\hat{j}+\hat{k})$,$B (\hat{i}-3 \hat{j}-5 \hat{k})$,and $C (3 \hat{i}-4 \hat{j}-4 \hat{k})$ are the vertices of a right-angled triangle.

The constant value $(\lambda + \mu)$ for which the lines $\vec{r} = (2\hat{i} + \hat{j} + \hat{k}) + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = (\hat{i} + \hat{j} - 3\hat{k}) + \mu(\hat{j} + 2\hat{k})$ intersect each other is equal to (where $\lambda$ and $\mu$ are parameters).

If the position vectors of $A, B$ and $C$ are respectively $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$,then $\cos ^2 A$ is equal to

If the vectors $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$ are orthogonal to each other,then the locus of the point $(x, y)$ is