If $a^2 + b^2 + c^2 = 1$,then the maximum possible value of $3a + 4b + 12c$ is equal to (where $a, b, c \in \mathbb{R}$)-

For all real $x$,the vectors $Cx \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2Cx \hat{k}$ make an obtuse angle with each other. Then the value of $C$ can be in:

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

If $\theta$ is the angle between the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $a \hat{i}+4 \hat{j}+b \hat{k}$ and $\cos \theta=\frac{2}{3}$,then $2(a+b+3)=$

If $\lambda > 0$,let $\theta$ be the angle between the vectors $\vec{a} = \hat{i} + \lambda \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. If the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular,then the value of $(14 \cos \theta)^2$ is equal to

If $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$ are such that the angle between the lines $DC$ and $AB$ is $\theta$,then $\cos \theta$ is equal to