If $a$ and $b$ are two non-zero perpendicular vectors,then a vector $y$ satisfying the equations $a \cdot y = c$ (where $c$ is a scalar) and $a \times y = b$ is

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is :

If $\vec{a}_1$ is the component of vector $\vec{a}$ along the direction of vector $\vec{b}$,and $\vec{a}_2$ is the component of $\vec{a}$ perpendicular to $\vec{b}$,then $\vec{a}_1 \times \vec{a}_2 = \dots$

Let $O$ be the origin,and $\overline{OX}, \overline{OY}, \overline{OZ}$ be three unit vectors in the directions of the sides $QR, RP, PQ$,respectively,of a triangle $PQR$.
$(1)$ Find $|\overline{OX} \times \overline{OY}|$.
$[A] \sin(P+Q)$
$[B] \sin 2R$
$[C] \sin(P+R)$
$[D] \sin(Q+R)$
$(2)$ If the triangle $PQR$ varies,then find the minimum value of $\cos(P+Q) + \cos(Q+R) + \cos(R+P)$.
$[A] -\frac{5}{3}$
$[B] -\frac{3}{2}$
$[C] \frac{3}{2}$
$[D] \frac{5}{3}$
Select the correct options for $(1)$ and $(2)$.

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$, then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to: