If the shortest distance between the lines $L_1:\vec{r}=(2+\lambda)\hat{i}+(1-3\lambda)\hat{j}+(3+4\lambda)\hat{k}$, $\lambda\in\mathbb{R}$ and $L_2:\vec{r}=2(1+\mu)\hat{i}+3(1+\mu)\hat{j}+(5+\mu)\hat{k}$, $\mu\in\mathbb{R}$ is $\dfrac{m}{\sqrt{n}}$, where $\gcd(m,n)=1$, then the value of $m+n$ equals:
Let $P(\alpha,\beta,\gamma)$ be the image of the point $Q(3,-3,1)$ in the line $\dfrac{x-0}{1}=\dfrac{y-3}{1}=\dfrac{z-1}{-1}$ and $R$ be the point $(2,5,-1)$. If the area of triangle $PQR$ is $\lambda$ and $\lambda^2=14K$, then $K$ is equal to:
Let the position vectors of the vertices $A$, $B$ and $C$ of a triangle be $2\hat{i}+2\hat{j}+\hat{k}$, $\hat{i}+2\hat{j}+2\hat{k}$ and $2\hat{i}+\hat{j}+2\hat{k}$ respectively. Let $l_1,l_2$ and $l_3$ be the lengths of perpendiculars drawn from the orthocentre of the triangle on the sides $AB$, $BC$ and $CA$ respectively, then $l_1^2+l_2^2+l_3^2$ equals:
Given a tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(−1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is: