If the vectors $\bar{a}, \bar{b}, \bar{c}$ satisfy the condition $|\bar{a}-\bar{c}|=|\bar{b}-\bar{c}|$,then $(\bar{b}-\bar{a}) \cdot \left(\bar{c}-\frac{\bar{a}+\bar{b}}{2}\right) = $

If $|a| = 3, |b| = 1, |c| = 4$ and $a + b + c = 0,$ then $a \cdot b + b \cdot c + c \cdot a = $

If the angle $\theta$ between the vectors $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ is such that $90^{\circ} < \theta < 180^{\circ}$,then $x$ lies in the interval

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}-\bar{b}|^2+|\bar{b}-\bar{c}|^2+|\bar{c}-\bar{a}|^2=15$,then $|\bar{a}-\bar{b}-\bar{c}|^2-4(\bar{b} \cdot \bar{c})=$

Let $A = \hat{i} + 2 \hat{j}$. If $B$ is a vector in the $XY$ plane such that $(A + B) \cdot B = 15$ and $A \cdot B = 6$,then $|B|$ is

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are the position vectors of the points $A, B, C, D$ respectively such that $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}=\overline{0}$,then the position vector of the point of intersection of the line segments $AC$ and $BD$ is