$A$ plane $\pi$ is passing through the points $A(1, -2, 3)$ and $B(6, 4, 5)$. If the plane $\pi$ is perpendicular to the plane $3x - y + z = 2$,then the perpendicular distance from $(0, 0, 0)$ to the plane $\pi$ is

  • A
    $\frac{63}{\sqrt{594}}$
  • B
    $\frac{32}{\sqrt{594}}$
  • C
    $\frac{72}{\sqrt{435}}$
  • D
    $\frac{23}{\sqrt{135}}$

Explore More

Similar Questions

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

The equation of the plane containing the line $r = i + j + \lambda (2i + j + 4k)$ is

Let $P$ be the plane $\sqrt{3} x+2 y+3 z=16$ and let $S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1 \text{ and the distance of } (\alpha, \beta, \gamma) \text{ from the plane } P \text{ is } \frac{7}{2}\right\}$. Let $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$ be three distinct vectors in $S$ such that $|\overrightarrow{u}-\overrightarrow{v}|=|\overrightarrow{v}-\overrightarrow{w}|=|\overrightarrow{w}-\overrightarrow{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ 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:

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

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo