The three points $A(2, 4, 3)$, $B(4, a, 9)$ and $C(10, -1, 7)$ form a right-angled triangle with $\angle B = 90^\circ$. The value of $a$ is

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively,are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

If $ABCD$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(AB)^2+(CD)^2=4R^2$,then:

In quadrilateral $ABCD$,$AB=a$,$BC=b$,$AD=b-a$. If $M$ is the midpoint of $BC$ and $N$ is a point on $DM$ such that $DN=\left(\frac{4}{5}\right) DM$,then $5 AN=$

$a, b$ and $c$ are three vectors with magnitudes $|a| = 4, |b| = 4, |c| = 2$ such that $a$ is perpendicular to $(b + c)$,$b$ is perpendicular to $(c + a)$,and $c$ is perpendicular to $(a + b)$. It follows that $|a + b + c|$ is equal to

Let $\hat{a}$ and $\hat{b}$ be two unit vectors. If the vectors $\vec{c} = \hat{a} + 2\hat{b}$ and $\vec{d} = 5\hat{a} - 4\hat{b}$ are perpendicular to each other,then the angle between $\hat{a}$ and $\hat{b}$ is: