If the four points,whose position vectors are $3 \hat{i} - 4 \hat{j} + 2 \hat{k}$,$\hat{i} + 2 \hat{j} - \hat{k}$,$-2 \hat{i} - \hat{j} + 3 \hat{k}$,and $5 \hat{i} - 2 \alpha \hat{j} + 4 \hat{k}$ are coplanar,then $\alpha$ is equal to

If $a, b$ and $c$ are three non-coplanar vectors and $p, q$ and $r$ are vectors defined by $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$,then $(a+b) \cdot p+(b+c) \cdot q+(c+a) \cdot r$ is

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector, let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$, $\vec{b} = \lambda \hat{i}+3 \hat{j}$, and we consider the standard basis vectors or a third vector to define coplanarity. However, if the question implies these two vectors are coplanar with the origin or a specific plane, we evaluate the scalar triple product. Given the standard interpretation of such problems, if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$, $\vec{b} = (\lambda, 3, 0)$, and $\vec{c} = (0, 1, 0)$:

Unit vectors $a, b, c$ are coplanar. $A$ unit vector $d$ is perpendicular to the given vectors. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^{\circ}$,then $c = ....$

$(a+b) \cdot(b+c) \times(a+b+c)$ is equal to

$(\vec{a}+2 \vec{b}-\vec{c}) \cdot \{(\vec{a}-\vec{b}) \times (\vec{a}-\vec{b}-\vec{c})\} =$