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 $a=2 \hat{i}-2 \hat{j}+\hat{k}$ and $b=-\hat{j}+\hat{k}$. If $c$ is a vector such that $a \cdot c=|c|$, $|c-a|=2 \sqrt{2}$, and the angle between $a \times b$ and $c$ is $\frac{\pi}{3}$, then $|(a \times b) \times c|=$

Let $A(2, 3, 0)$, $B(0, 3, 2)$ and $C(4, 0, 3)$ be the vertices of a triangle. Find the area of the triangle.

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two perpendicular lines,then the direction cosines of the line which is perpendicular to both the lines will be:

Let $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$,$\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c}=-3$. Then $|\overrightarrow{c}|^2$ is equal to . . . . . . .

The area of the parallelogram whose diagonals are represented by the vectors $\bar{a}=3 \hat{i}-\hat{j}-2 \hat{k}$ and $\bar{b}=-\hat{i}+3 \hat{j}-3 \hat{k}$ is