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:
The sum of the intercepts on the coordinate axes of the plane passing through the point (–2, –2, 2) and containing the line joining the points (1, –1, 2) and (1, 1, 1), is
Let P(x, y, z) be any point on the locus, then the distances from the six faces are |x + 1|, |x − 1|, |y + 1|, |y − 1|, |z + 1| and |z − 1|. According to the given condition, find the locus of P.\(|x+1|^2 + |x-1|^2 + |y+1|^2 + |y-1|^2 + |z+1|^2 + |z-1|^2 = 10\)