Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 3: Exercise-3

If the distance of the point $P\left(1,-2,1\right)$ from the plane $x+2y-2z=\alpha$, where $\alpha >0,$ is $5$ then the foot of the perpendicular from $P$ to the plane is

Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x-y+z=3.$ Then, the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

Let $P$ be a point in the first octant, whose image $Q$ in the plane $x+y=3$ (that is, the line segment $PQ$ is perpendicular to the plane $x+y=3$ and the mid-point of $PQ$ lies in the plane $x+y=3$) lies on the $Z-$axis. Let the distance of $P$ from the $X-$axis be $5.$ If $R$ is the image of $P$ in the $XY-$plane, then the length of $PR$ is........

The distance of the point $\left(1,0,2\right)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=16,$ is 

If the image of the point $P\left(1,-2,3\right)$ in the plane, $2x+3y-4z+22=0$ measured parallel to the line, $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $Q$, then $PQ$ is equal to

The length of the projection of the line segment joining the points $\left(5,-1,4\right)$ and $\left(4,-1,3\right)$ on the plane, $x+y+z=7$ is

If the lines $x=a y+b, z=c y+d$ and $x=a^{\text{'}}z+b^{\text{'}},y=c^{\text{'}}z+d^{\text{'}}$ are perpendicular, then :

If $L_1$ is the line of intersection of the planes $2x-2y+3z-2=0, x-y+z+1=0$ and $L_2$ is the line of intersection of the planes $x+2y-z-3=0, 3x-y+2z-1=0,$ then the distance of the origin from the plane containing the lines $L_1$ and $L_2$ is