The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is

  • A
    $7$
  • B
    $1$
  • C
    $\frac{1}{7}$
  • D
    $\frac{7}{5}$

Explore More

Similar Questions

The equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$ is:

The largest value of $a$,for which the perpendicular distance of the plane containing the lines $\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a\hat{j}-\hat{k})$ and $\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a\hat{k})$ from the point $(2,1,4)$ is $\sqrt{3}$,is...

The foot of the perpendicular from a point on the circle $x^{2} + y^{2} = 1, z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves?

The line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point:

$L$ is a line passing through the point $A(1, 0, -3)$ and parallel to a line having direction ratios $0, 1, -2$. $P$ is a point on the line $L$ which is at a minimum distance from the plane $2x + 3y + 5z = 1$. Then, the equation of the plane through $P$ and perpendicular to $AP$ 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