Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

  • A
    $5x - 11y + z = 17$
  • B
    $\sqrt{2}x + y = 3\sqrt{2} - 1$
  • C
    $x + y + z = \sqrt{3}$
  • D
    $x - \sqrt{2}y = 1 - \sqrt{2}$

Explore More

Similar Questions

If the equation of the plane containing the line $x+2y+3z-4=0=2x+y-z+5$ and perpendicular to the plane $\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2\hat{j}+3\hat{k})$ is $ax+by+cz=4$,then $(a-b+c)$ is equal to

The position vectors of points $A$ and $B$ are $i - j + 3k$ and $3i + 3j + 3k$ respectively. The equation of a plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. The points $A$ and $B$:

The lines $\frac{x-0}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda}$ and $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{\lambda}$ are coplanar. If $p$ is the plane containing these lines,then which of the following points lies on the plane for all values of $\lambda$?

The plane $lx + my = 0$ is rotated by an angle $\alpha$ about its line of intersection with the plane $z = 0$. Find the equation of the plane in its new position.

Difficult
View Solution

$A$ plane $\pi_1$ contains the vectors $\bar{i}+\bar{j}$ and $\bar{i}+2\bar{j}$. Another plane $\pi_2$ contains the vectors $2\bar{i}-\bar{j}$ and $3\bar{i}+2\bar{k}$. $\bar{a}$ is a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\bar{a}$ and $\bar{i}-2\bar{j}+2\bar{k}$ is acute, then $\theta=$

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