Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: KCET (UG) 2020

The mid-points of the sides of triangle are $\left(1, 5,-1\right), \left(0, 4,-2\right)$ and $\left(2,3,4\right)$, then centroid of the triangle is

The equation of the line joining the points $(\left(-3, 4, 11\right)$ and $(1, -2, 7)$ is

The image of the point $\left(1,6,3\right)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is

The angle between the lines $2x=3y=-z$ and $6x=-y=-4z$ is

The angle between the lines whose direction cosines are $\left(\frac{\sqrt{3}}{4},\frac{1}{4},\frac{\sqrt{3}}{2}\right)$ and $\left(\frac{\sqrt{3}}{4},\frac{1}{4},\frac{-\sqrt{3}}{2}\right)$ is

If a plane meets the coordinate axes at $A, B$ and $C$ in such a way that the centroid of triangle $ABC$ is at the point $\left(1, 2, 3\right)$ then the equation of the plane is

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$ is

The locus represented by $xy+yz=0$ is