If $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$,then $a$ is equal to

If $\lambda > 0$,let $\theta$ be the angle between the vectors $\vec{a} = \hat{i} + \lambda \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. If the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular,then the value of $(14 \cos \theta)^2$ is equal to

The three points $A(2, 4, 3)$, $B(4, a, 9)$ and $C(10, -1, 7)$ form a right-angled triangle with $\angle B = 90^\circ$. The value of $a$ is

If $\theta$ is the angle between the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $a \hat{i}+4 \hat{j}+b \hat{k}$ and $\cos \theta=\frac{2}{3}$, then $2(a+b+3)=$

Let $\bar{a} = \bar{i} + 2\bar{j} + 3\bar{k}$, $\bar{b} = 2\bar{i} - 3\bar{j} + \bar{k}$, and $\bar{c} = 3\bar{i} + \bar{j} - 2\bar{k}$ be three vectors. If $\bar{r}$ is a vector such that $\bar{r} \cdot \bar{a} = 0$, $\bar{r} \cdot \bar{b} = -2$, and $\bar{r} \cdot \bar{c} = 6$, then find the value of $\bar{r} \cdot (3\bar{i} + \bar{j} + \bar{k})$.

If the projection of $\bar{a}$ on $\bar{b}+\bar{c}$ is twice the projection of $\bar{b}+\bar{c}$ on $\bar{a}$,and if $|\bar{b}|=2 \sqrt{2}$,$|\bar{c}|=4$,and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$,then $|\bar{a}|=$