Let $a$,$b$,and $c$ be real numbers such that $a^2 + b^2 + c^2 = 1$. Then the maximum value of $(4b - 3c)^2 + (4a - 2c)^2 + (3a - 2b)^2$ is:

Let $\overrightarrow{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=((\overrightarrow{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

Let $p, q, r$ be three mutually perpendicular vectors of the same magnitude. If a vector $x$ satisfies the equation $p \times \{(x - q) \times p\} + q \times \{(x - r) \times q\} + r \times \{(x - p) \times r\} = 0$,then $x$ is given by

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-2\hat{k}, \vec{c}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{d}=-4\hat{i}+5\hat{j}-3\hat{k}$. If $\vec{d}=x(\vec{b} \times \vec{c})-\frac{7}{9}(\vec{c} \times \vec{a})+z(\vec{a} \times \vec{b})$,then find the value of $x$.

If $|\overline{a}|=3$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$,the angle between $\overline{b}$ and $\overline{c}$ is $\frac{\pi}{3}$,and $\overline{a}$ is perpendicular to $\overline{b} \times \overline{c}$,then the value of $|\overline{a} \times(\overline{b} \times \overline{c})|$ is

Find the unit vector which is perpendicular to the vector $5i + 2j + 6k$ and coplanar with the vectors $2i + j + k$ and $i - j + k$.