If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of points $A, B, C$ respectively,and $D$ is the midpoint of $BC$,then $\vec{AD} = \dots$

The direction cosines of the vector $\vec{a} = -2 \hat{i} + \hat{j} - 5 \hat{k}$ are

$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = $

If $a = (2, 5)$ and $b = (1, 4),$ then the vector parallel to $(a + b)$ is

The position vectors of the points $A, B, C$ are $(2i + j - k)$,$(3i - 2j + k)$,and $(i + 4j - 3k)$ respectively. These points

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 $17(\lambda + 9) =$ ?