Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$, whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

  • A
    $34$
  • B
    $28$
  • C
    $40$
  • D
    $6$

Explore More

Similar Questions

If the lines $\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$ intersect each other,then the integer value of $k$ is:

Difficult
View Solution

Let $l_1$ be the line passing through the point $A = 3\hat{i} + 4\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_1} = -\hat{i} + 2\hat{j} + \hat{k}$. Let $l_2$ be another line passing through the point $B = \hat{i} - 7\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$. Then the shortest distance between the lines $l_1$ and $l_2$ is:

The shortest distance between lines $\overline{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\overline{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$ is

$A$ line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$,where $a, b, c$ are non-negative integers,lies on the line $AB$ at a distance of $21$ units from point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to...........

The equation of the line,passing through $A(1, 2, 3)$ and perpendicular to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} + 3 \hat{j} + 2 \hat{k}$,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