Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{c}=\hat{i}-\hat{j}$ and $\vec{d}=\hat{i}+\hat{j}+x \hat{k}$. If $(\vec{a} \times \vec{b}) \times \vec{c}$ is perpendicular to $\vec{d}$, then $x=$

Let $\overrightarrow{a}=\hat{i}+5\hat{j}+\alpha\hat{k}$,$\overrightarrow{b}=\hat{i}+3\hat{j}+\beta\hat{k}$ and $\overrightarrow{c}=-\hat{i}+2\hat{j}-3\hat{k}$ be three vectors such that $|\overrightarrow{b} \times \overrightarrow{c}|=5\sqrt{3}$ and $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$. Then the greatest value of $|\vec{a}|^{2}$ is .... .

Let $\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}$,$\bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}$,$\bar{c}=3 \bar{i}+\bar{j}-\bar{k}$ and let $\bar{\alpha}$ be a vector perpendicular to both $\bar{a}$ and $\bar{b}$ such that $\bar{\alpha} \cdot \bar{c}=63$. Then $\bar{\alpha}=$

The points $A(a), B(b), C(c)$ will be collinear if

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

If the two diagonals of a parallelogram are $\bar{d_1} = \bar{i} + 2\bar{j} + 3\bar{k}$ and $\bar{d_2} = -2\bar{i} + \bar{j} - 2\bar{k}$,then the area of the parallelogram in square units is