The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
Let $R$ be a relation defined on the set $\{1,2,3,4\times\{1,2,3,4\}$ by \[ R=\{((a,b),(c,d)) : 2a+3b=3c+4d\} \] Then the number of elements in $R$ is
Let $\vec a = 2\hat i + \hat j - 2\hat k$, $\vec b = \hat i + \hat j$ and $\vec c = \vec a \times \vec b$. Let $\vec d$ be a vector such that $|\vec d - \vec a| = \sqrt{11}$, $|\vec c \times \vec d| = 3$ and the angle between $\vec c$ and $\vec d$ is $\frac{\pi}{4}$. Then $\vec a \cdot \vec d$ is equal to
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$
Let \(S=\left\{ z\in\mathbb{C}:\left|\frac{z-6i}{z-2i}\right|=1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5} \right\}.\)Then $\sum_{z\in S}|z|^2$ is equal to