1
JEE Main 2023 (Online) 31st January Morning Shift
+4
-1

Let the shortest distance between the lines

$$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$$ and

$$L_{1}: x+1=y-1=4-z$$ be $$2 \sqrt{6}$$. If $$(\alpha, \beta, \gamma)$$ lies on $$L$$,

then which of the following is NOT possible?

A
$$\alpha+2 \gamma=24$$
B
$$2 \alpha+\gamma=7$$
C
$$\alpha-2 \gamma=19$$
D
$$2 \alpha-\gamma=9$$
2
JEE Main 2023 (Online) 30th January Evening Shift
+4
-1
Out of Syllabus
A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then :
A
$a+b+\sqrt{2} c=1$
B
$\sqrt{2} a+b+c=1$
C
$\sqrt{2} a-b+c=1$
D
$a+\sqrt{2} b+c=1$
3
JEE Main 2023 (Online) 30th January Evening Shift
+4
-1
Out of Syllabus
If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is :
A
$\frac{17}{5}$
B
$\frac{6}{13}$
C
$\frac{13}{6}$
D
$\frac{5}{17}$
4
JEE Main 2023 (Online) 30th January Morning Shift
+4
-1
Out of Syllabus

The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :

A
9
B
7
C
$$\frac{19}{3}$$
D
$$\frac{13}{3}$$
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