1
JEE Main 2024 (Online) 1st February Morning Shift
+4
-1
If the shortest distance between the lines

$\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\lambda$ is :
A
0
B
$2 \sqrt{3}$
C
$3 \sqrt{3}$
D
$-2 \sqrt{3}$
2
JEE Main 2024 (Online) 31st January Evening Shift
+4
-1

Let $$(\alpha, \beta, \gamma)$$ be the mirror image of the point $$(2,3,5)$$ in the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$. Then, $$2 \alpha+3 \beta+4 \gamma$$ is equal to

A
32
B
31
C
33
D
34
3
JEE Main 2024 (Online) 31st January Evening Shift
+4
-1

The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is

A
$$\frac{141}{\sqrt{221}}$$
B
$$\frac{24}{\sqrt{117}}$$
C
$$\frac{42}{\sqrt{117}}$$
D
$$\frac{121}{\sqrt{221}}$$
4
JEE Main 2024 (Online) 30th January Evening Shift
+4
-1

Let $$L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$$,

$$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$$

be three lines such that $$L_1$$ is perpendicular to $$L_2$$ and $$L_3$$ is perpendicular to both $$L_1$$ and $$L_2$$. Then, the point which lies on $$L_3$$ is

A
$$(1,7,-4)$$
B
$$(1,-7,4)$$
C
$$(-1,7,4)$$
D
$$(-, 1-7,4)$$
EXAM MAP
Medical
NEET