Three-Dimensional Geometry

The distance of the point (1, −5, 9) from the plane $x-y+z=5$ measured along the line $x=y=z$ is :

If the line, $\dfrac{x-3}2=\dfrac{y+2}{-1}=\dfrac{z+4}3$ lies in the plane, $lx+my-z=9$, then $l^2+m^2$ is equal to :

The shortest distance between the lines $\dfrac x2=\dfrac y2=\dfrac z1$ and $\dfrac{x+2}{-1}=\dfrac{y-4}8=\dfrac{z-5}4$ lies in the interval :

The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$, is :

ABC is a triangle in a plane with vertices A(2, 3, 5), B(−1, 3, 2) and C(λ, 5, μ). If the median through A is equally inclined to the coordinate axes, then the value of $(\lambda^3+\mu^3+5)$ is :

The number of distinct real values of λ for which the lines $\dfrac{x-1}1=\dfrac{y-2}2=\dfrac{z+3}{\lambda^2}$ and $\dfrac{x-3}1=\dfrac{y-2}{\lambda^2}=\dfrac{z-1}2$ are coplanar is :

In $\mathbb R^3$, consider the planes $P_1:y=0$ and $P_2:x+z=1$. Let $P_3$ be a plane, different from $P_1$ and $P_2$, which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0,1,0)$ from $P_3$ is 1 and the distance of a point $(\alpha,\beta,\gamma)$ from $P_3$ is 2, then which of the following relations is (are) true?

In $\mathbb R^3$, let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_1:x+2y-z+1=0$ and $P_2:2x-y+z-1=0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_1$. Which of the following points lie(s) on $M$?

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

The equation of the plane containing the line $2x-5y+z=3$; $x+y+4z=5$, and parallel to the plane, $x+3y+6z=1$, is :

If the points $(1,1,\lambda)$ and $(-3,0,1)$ are equidistant from the plane, $3x+4y-12z+13=0$, then λ satisfies the equation :

If the shortest distance between the lines $\dfrac{x-1}\alpha=\dfrac{y+1}{-1}=\dfrac z1$, $(\alpha\ne-1)$ and $x+y+z+1=0=2x-y+z+3$ is $\dfrac1{\sqrt3}$, then a value of α is :

The shortest distance between the $z$-axis and the line $x+y+2z-3=0=2x+3y+4z-4$, is :

A plane containing the point (3, 2, 0) and the line $\dfrac{x-1}1=\dfrac{y-2}5=\dfrac{z-3}4$ also contains the point :

From a point $P(\lambda,\lambda,\lambda)$, perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x,\ z=1$ and $y=-x,\ z=-1$. If $P$ is such that $\angle QPR$ is a right angle, then the possible value(s) of $\lambda$ is(are)

The image of the line $\dfrac{x-1}{3}=\dfrac{y-3}{1}=\dfrac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line :

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2=m^2+n^2$ is :

Equation of the plane which passes through the point of intersection of lines
$$\frac{x-1}3=\frac{y-2}1=\frac{z-3}2\quad\text{and}\quad\frac{x-3}1=\frac{y-1}2=\frac{z-2}3$$
and has the largest distance from the origin is :

A line in the 3-dimensional space makes an angle $\theta\left(0<\theta\le\dfrac\pi2\right)$ with both the $x$ and $y$ axes. Then the set of all values of $\theta$ is the interval :

Let $A(2,3,5)$, $B(-1,3,2)$ and $C(\lambda,5,\mu)$ be the vertices of a $\Delta ABC$. If the median through A is equally inclined to the coordinate axes, then :