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 $|\bar{a}|=$

Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$,then the cosine of the angle $\alpha$ is given by

Let $m$ be the unit vector orthogonal to the vector $\hat{i}-\hat{j}+\hat{k}$ and coplanar with the vectors $2 \hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. If $a=\hat{i}-\hat{k}$,then the length of the perpendicular from the origin to the plane $r \cdot m=a \cdot m$ is

If $\overline{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\overline{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\overline{c}=3 \hat{i}+\hat{j}$ such that $\overline{b}+\lambda \overline{a}$ is perpendicular to $\overline{c}$,then $\lambda$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that the angle between any two of them is the same. If the product of their magnitudes is $14$ and $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168$,then $|\vec{a}| + |\vec{b}| + |\vec{c}|$ is equal to:

Let a vector $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$, $\lambda>0$, make an obtuse angle with the vector $\overrightarrow{b}=-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$ and an angle $\theta$, $\frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to . . . . . . .