1
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$.
Let the line $\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is equal to :

A

25

B

27

C

19

D

29

2
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines.

Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $\mathrm{L}_1$, then $|5 \alpha-11 \beta-8 \gamma|$ equals :

A

25

B

20

C

16

D

18

3
JEE Main 2025 (Online) 28th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The square of the distance of the point $ \left( \frac{15}{7}, \frac{32}{7}, 7 \right) $ from the line $ \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} $ in the direction of the vector $ \hat{i} + 4\hat{j} + 7\hat{k} $ is:

A

66

B

54

C

41

D

44

4
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\mathrm{A}(x, y, z)$ be a point in $x y$-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and $(0,0,1)$.

Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements

(S1) : $\triangle \mathrm{ABC}$ is an isosceles right angled triangle, and

(S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$,

A
both are false
B
only (S2) is true
C
only (S1) is true
D
both are true
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