The vectors $a$ and $b$ are non-collinear. The value of $x$ for which the vectors $c = (x - 2)a + b$ and $d = (2x + 1)a - b$ are collinear,is

If $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$,$b = 2 \hat{i} + 3 \hat{j} + \hat{k}$,$c = 8 \hat{i} + 13 \hat{j} + 9 \hat{k}$ and $x a + y b + z c = 0$,then $\frac{x y}{z^2} =$

The position vector of the point of intersection of the medians (centroid) of a triangle,whose vertices are $A(1, 2, 3)$,$B(1, 0, 3)$,and $C(4, 1, -3)$ is

If the vectors $\overline{AB}=3 \hat{i}+4 \hat{k}$ and $\overline{AC}=5 \hat{i}-2 \hat{j}+4 \hat{k}$ are the sides of the triangle $ABC$,then the length of the median through $A$ is:

For any vector $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$,with $10|a_i| < 1$,$i = 1, 2, 3$,consider the following statements:
$(A): \max \{|a_1|, |a_2|, |a_3|\} \leq |\vec{a}|$
$(B): |\vec{a}| \leq 3 \max \{|a_1|, |a_2|, |a_3|\}$

If $A, B, C, D$ are any four points and $E$ and $F$ are the midpoints of $AC$ and $BD$ respectively,then $\overline{AB} + \overline{CB} + \overline{CD} + \overline{AD} = \dots$