1
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Let the image of the point $$\mathrm{P}(1,2,6)$$ in the plane passing through the points $$\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$$ and $$\mathrm{C}(0,5,1)$$ be $$\mathrm{Q}(\alpha, \beta, \gamma)$$. Then $$\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$$ is equal to :

A
76
B
62
C
70
D
65
2
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\mathrm{AB}$$ from the plane $$2 x-2 y+z=14$$ is :

A
3
B
$$\frac{10}{3}$$
C
4
D
$$\frac{11}{3}$$
3
JEE Main 2023 (Online) 10th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The shortest distance between the lines $${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$$ and $${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$$ is :

A
8
B
7
C
6
D
9
4
JEE Main 2023 (Online) 10th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\alpha,\beta,\gamma)$$, then $$\alpha\beta+\beta\gamma+\gamma\alpha$$ is equal to :

A
72
B
74
C
76
D
70
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