The angle between the diagonals of the parallelogram whose adjacent sides are $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\hat{i}+2 \hat{j}+3 \hat{k}$ is

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}-\bar{b}|^2+|\bar{b}-\bar{c}|^2+|\bar{c}-\bar{a}|^2=15$,then $|\bar{a}-\bar{b}-\bar{c}|^2-4(\bar{b} \cdot \bar{c})=$

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ with $a_{i} > 0$ for $i = 1, 2, 3$ be a vector that makes equal angles with the coordinate axes $OX$,$OY$,and $OZ$. Also,let the projection of $\vec{a}$ on the vector $3 \hat{i} + 4 \hat{j}$ be $7$. Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ by $90^{\circ}$. If $\vec{a}$,$\vec{b}$,and the $x$-axis are coplanar,then the projection of vector $\vec{b}$ on $3 \hat{i} + 4 \hat{j}$ is equal to

If $a, b, c$ are the position vectors of the points $A, B, C$ respectively, then match the items of List-$I$ with those of List-$II$.

List-$I$	List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$	$I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$	$II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$	$III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$	$IV$. $A, B, C$ are collinear

The correct match is:

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$

Let $|\vec{a}| = |\vec{b}| = 1$ and $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c}$ is a vector satisfying the condition $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} \times \vec{b})$,then $\vec{c} \cdot \vec{b} = \dots$ (in $/2$)