Let a line with direction ratios $a, -4a, -7$ be perpendicular to the lines with direction ratios $3, -1, 2b$ and $b, a, -2$. If the point of intersection of the line $\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$ and the plane $x - y + z = 0$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to $.......$

  • A
    $20$
  • B
    $10$
  • C
    $30$
  • D
    $40$

Explore More

Similar Questions

Find the vector equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0$.

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2x + ky - 5z = 1$ and $3kx - ky + z = 5$,where $k < 3$,and intercepts a unit length on the positive $x$-axis,then the intercept made by the plane $P$ on the $y$-axis is

The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

The vector equation of any plane passing through the line of intersection of the planes $\vec{r} \cdot \vec{m}_1=q_1$ and $\vec{r} \cdot \vec{m}_2=q_2$ is given by $\vec{r} \cdot (\vec{m}_1+\lambda \vec{m}_2)=q_1+\lambda q_2$ for $\lambda \in R$. Find the vector equation of the plane passing through the point $2 \hat{i}-3 \hat{j}+\hat{k}$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=5$ and $\vec{r} \cdot (3 \hat{i}+\hat{j}-2 \hat{k})=7$.

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane $x + 2y + 3z = 15$ at a point $P$,then the distance of $P$ from the origin 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