The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors $\bar{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\bar{c}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}$ is $1$. Then the value of $\lambda$ is:

If the vectors $\bar{a}, \bar{b}, \bar{c}$ satisfy the condition $|\bar{a}-\bar{c}|=|\bar{b}-\bar{c}|$,then $(\bar{b}-\bar{a}) \cdot \left(\bar{c}-\frac{\bar{a}+\bar{b}}{2}\right) = $

If the vector $a = 3\hat{j} + 4\hat{k}$ is the sum of two vectors $a_1$ and $a_2$, where vector $a_1$ is parallel to $b = \hat{i} + \hat{j}$ and vector $a_2$ is perpendicular to $b$, then $a_1 =$

Let a unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$ and $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\overrightarrow{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,then $|\hat{u}-\overrightarrow{v}|^2$ is equal to

If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\bar{a}|=5$, $|\bar{b}|=12$ and $|\bar{a}-\bar{b}|=13$, then $|2\bar{a}+\bar{b}|=$

$A$ parallelogram is constructed with $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$ as its adjacent sides, where $|\vec{a}| = 2\sqrt{2}$ and $|\vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. Find the lengths of the diagonals of the parallelogram.