If $\bar{a} = 2\bar{i} - 3\bar{j} + 5\bar{k}$ and $\bar{b} = -\bar{i} + 3\bar{j} + 3\bar{k}$ are two vectors,then the vector of magnitude $28$ units in the direction of the vector $\bar{a} - \bar{b}$ is

For any two points $M$ and $N$ in the $XY$-plane, let $\overrightarrow{MN}$ denote the vector from $M$ to $N$, and $\overrightarrow{0}$ denote the zero vector. Let $P, Q$ and $R$ be three distinct points in the $XY$-plane. Let $S$ be a point inside the triangle $\triangle PQR$ such that $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \overrightarrow{0}$. Let $E$ and $F$ be the mid-points of the sides $PR$ and $QR$, respectively. Then the value of $\frac{\text{length of the line segment } EF}{\text{length of the line segment } ES}$ is: (in $.20$)

$A$ vector coplanar with the non-collinear vectors $\vec{a}$ and $\vec{b}$ is:

If the position vectors of the points $A, B, C, D$ given by $\hat{i}+2 \hat{j}+3 \hat{k}, 2 \hat{i}-\hat{j}+2 \hat{k}$, $\frac{1}{4}(7 \hat{i}+15 \hat{j}+15 \hat{k})$ and $\frac{1}{3}[7 \hat{i}+2 \hat{j}+(5+3 a) \hat{k}]$ respectively are such that $|AC|=|BD|$, then $16(3a-1)^2=$

Five points given by $A, B, C, D, E$ are in a plane. Three forces $\overrightarrow{AC}, \overrightarrow{AD},$ and $\overrightarrow{AE}$ act at $A$,and three forces $\overrightarrow{CB}, \overrightarrow{DB},$ and $\overrightarrow{EB}$ act at $B$. Then their resultant is:

If $A(2 \hat{i} + \hat{j} - \hat{k})$, $B(\lambda \hat{i} + 5 \hat{j} + 4 \hat{k})$, $C(-4 \hat{i} + 3 \hat{j} + 2 \hat{k})$ and $D(-\hat{i} - 2 \hat{j} + 3 \hat{k})$ are four points in space such that $\overrightarrow{AB} = x \overrightarrow{AC} + y \overrightarrow{AD}$ for some real numbers $x \neq 0, y \neq 0$, then $17(\lambda + 9) =$ ?