$A$ unit vector $\vec{e} = a \hat{i} + b \hat{j} + c \hat{k}$ is coplanar with the vectors $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $3 \hat{i} + \hat{j} - 5 \hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$, then $2 a^2 + 3 b^2 + 4 c^2 =$

If $\overline{p}=\hat{i}+\hat{j}+\hat{k}$ and $\overline{q}=\hat{i}-2 \hat{j}+\hat{k}$. Then a vector of magnitude $5 \sqrt{3}$ units perpendicular to the vector $\overline{q}$ and coplanar with $\overline{p}$ and $\overline{q}$ is

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to:

$[\hat{i}-\hat{j}, \hat{j}-\hat{k}, \hat{k}-\hat{i}]$ is equal to

Let $a = i - k$, $b = xi + j + (1 - x)k$, and $c = yi + xj + (1 + x - y)k$. Then $[a\,b\,c]$ depends on

The value of $\lambda$ for which the four points $2i + 3j - k$,$i + 2j + 3k$,$3i + 4j - 2k$,and $i - \lambda j + 6k$ are coplanar is: