3d planes practice · Mixed difficulty

The line $\frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}$ and the plane $4 x+5 y+3 z-5=0$ intersect at a point

The point equidistant from the points $(a, 0,0),(0, b, 0),(0,0, c)$ and $(0,0,0)$ is

If a variable plane forms a tetrahedron of constant volume $64 \mathrm{k}^3$ with the coordinate planes then the locus of the centroid of the tetrahedron is $\mathrm{xyz}=\mathrm{uk}^3$. Find $\mathrm{u}$.

The distance of the point (1, 0, 2) 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

The plane through the intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to Y-axis also passes through the point

The distance of the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$ from the point $(-1,-5,-10)$ is

Assertion: Each part in which a line divides the cartesian plane, is known as half plane.

Reason: A point in the cartesian plane will either lie on a line or will lie in either of half plane I or II.

The equation of the plane which contains the y-axis and passes through the point (1,2,3) is

The plane containing the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and parallel to the line $\frac{x}{1}=\frac{y}{1}=\frac{z}{4}$ passes through the point:

The equation of the plane through the point $(2,5,-3)$ perpendicular to the planes $x+2 y+2 z=1$ and $x-2 y+3 z=4$ is -

The point of intersection of the line passing through $(0,0,1)$ and intersecting the lines $x+2 y+z=1$, $-x+y-2 z=2$ and $x+y=2, x+z=2$ with $x y$ plane is

The $X-Y$ plane is rotated about it's line of intersection with $Y-Z$ plane by $45^{\circ}$, then the equation of new plane is

A variable plane makes intercepts on $X, Y$ and $Z$-axes and it makes a tetrahedron of volume $64 \mathrm{cu}$. u. The locus of foot of perpendicular from origin on this plane is

If the plane $3 x+y+2 z+6-0$ is parallel to the line $\frac{3 x-1}{2 b}=3-y=\frac{z-1}{a}$, then the value of $3 a+3 b$ is

The length of the perpendicular from the point (2, -1, 4) on the straight line, $\frac{{x + 3}}{{10}} = \frac{{y - 2}}{{ - 7}} = \frac{z}{1}\,$ is

If from a point $P(a, b, c)$ perpendiculars $P A$ and $P B$ are drawn to $y z$ and $z x$ planes, then the equation of the plane $O A B$ is

In the space the equation $b y+c z+d=0$ represents a plane perpendicular to the plane

If the three planes $x=5,2 x-5 a y+3 z-2=0$ and $3 b x+y-3 z=0$ contain a common line, then $(a, b)$ is equal to

If the planes $x-y-z=0, x-y+\sin \alpha \cdot z=0$ and $x+\sin \alpha \cdot y-z=0$ pass through a straight line, then number of values of $\alpha$ in $[0,2021 \pi]$ is/are

The direction ratios of normal to the plane through the points (0, - 1, 0) and (0, 0, 1) and making an angle $\frac{\pi }{4}$ with the plane y – z + 5 = 0 are

Let $L$ be the line of intersection of the planes $2 x+3 y+z=1$ and $x+3 y+2 z=2$. If $L$ makes an angle $\alpha$ with the positive $X$-axis, then $\cos \alpha$ equals

An equation of a plane parallel to the plane $x-2 y+2 z-5=0$ and at a unit distance from the origin is

A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2 x+y+z=9$ at point $Q$, then the length $P Q$ equals

Equation of plane containing $(1,1,2)$ and which is at a maximum distance from $(2,2,1)$ is

The vector equation of the plane through the point $(2,1,-1)$ and passing through the line of intersection of the plane $\mathbf{r} \cdot(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}})=0$ and $\mathbf{r} \cdot(\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0$, is

The shortest distance from $(1,1,1)$ to the line of intersection of the pair of planes $x y+y z+z x+y^2=0$ is

Let $N$ be the foot of perpendicular from the point $P(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$. Then the distance of $N$ from the plane $2 x-2 y+z+5=0$ is

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

The d.r. of normal to the plane through $(1,0,0),(0,1,0)$ which makes an angle $\pi / 4$ with plane $x+y=3$ are

The equation of the plane through the origin containing the line $\frac{x-1}{5}=\frac{y-2}{4}=\frac{z-3}{5}$ is