If the position vectors of the vertices of a triangle are $2i + 4j - k,$ $4i + 5j + k,$ and $3i + 6j - 3k,$ then the triangle is

If $P=(0,1,2)$, $Q=(4,-2,1)$, and $O=(0,0,0)$, then $\angle POQ$ is equal to

The projection of the vector $2i + j - 3k$ on the vector $i - 2j + k$ is:

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for $K$ equal to:

$OA$ and $OB$ are two vectors of magnitudes $5$ and $6$ respectively. If $\angle BOA = 60^{\circ}$,then $OA \cdot OB$ is equal to

Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$ with the vector $\vec{a}+\vec{b}+\vec{c}$. Then $36 \cos ^{2} 2 \theta$ is equal to $.....$