The Cartesian equation of the plane passing through the point $(3, -2, -1)$ and parallel to the vectors $\vec{b} = \hat{i} - 2\hat{j} + 4\hat{k}$ and $\vec{c} = 3\hat{i} + 2\hat{j} - 5\hat{k}$ is:

  • A
    $2x - 17y - 8z + 63 = 0$
  • B
    $3x + 17y + 8z - 36 = 0$
  • C
    $2x + 17y + 8z + 36 = 0$
  • D
    $3x - 16y + 8z - 63 = 0$

Explore More

Similar Questions

The distance of the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$ from the origin is

If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2x - 5y = 15$,then $2\alpha - 3\beta$ is equal to:

The equation of the plane passing through the point $(1,1,1)$ and perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$ is

The distance of the point $(5,3,-1)$ from the plane passing through the points $A(2,1,0)$,$B(3,-2,4)$,and $C(1,-3,3)$ is:

The length of the perpendicular from the origin to the plane passing through three non-collinear points $a, b, c$ 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