1
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

2
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
3
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

A
12
B
9
C
7
D
8
4
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let in a $\triangle A B C$, the length of the side $A C$ be 6 , the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle A B C$ is:

A
42
B
17
C
56
D
21
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