The angle between the vectors $(2i + 6j + 3k)$ and $(12i - 4j + 3k)$ is

Let three vectors $\overrightarrow{a}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}$,$\vec{b}=5 \hat{i}+3 \hat{j}+4 \hat{k}$,and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ form a triangle such that $\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number,then $|\overrightarrow{c}|^2$ is:

If $A, B, C$ and $D$ are points whose position vectors are $\hat{i}+\hat{j}+\hat{k}, 4 \hat{i}-\hat{j}+2 \hat{k}, 5 \hat{i}+\hat{j}$ and $7 \hat{i}+2 \hat{j}+3 \hat{k}$ respectively,then the projection of $\vec{AB}$ on $\vec{CD}$ is

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

If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $\frac{\pi}{3}$,then the value of $|\vec{a}+\vec{b}|$ is

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to :