1
JEE Main 2023 (Online) 8th April Evening Shift
+4
-1
Out of Syllabus

For $$\mathrm{a}, \mathrm{b} \in \mathbb{Z}$$ and $$|\mathrm{a}-\mathrm{b}| \leq 10$$, let the angle between the plane $$\mathrm{P}: \mathrm{ax}+y-\mathrm{z}=\mathrm{b}$$ and the line $$l: x-1=\mathrm{a}-y=z+1$$ be $$\cos ^{-1}\left(\frac{1}{3}\right)$$. If the distance of the point $$(6,-6,4)$$ from the plane P is $$3 \sqrt{6}$$, then $$a^{4}+b^{2}$$ is equal to :

A
48
B
85
C
32
D
25
2
JEE Main 2023 (Online) 8th April Evening Shift
+4
-1
Out of Syllabus

Let $$\mathrm{P}$$ be the plane passing through the line

$$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$$ and the point $$(2,4,-3)$$.

If the image of the point $$(-1,3,4)$$ in the plane P

is $$(\alpha, \beta, \gamma)$$ then $$\alpha+\beta+\gamma$$ is equal to :

A
10
B
12
C
9
D
11
3
JEE Main 2023 (Online) 8th April Morning Shift
+4
-1

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

A
$$3 \sqrt{6}$$
B
$$6 \sqrt{2}$$
C
$$6 \sqrt{3}$$
D
$$2 \sqrt{6}$$
4
JEE Main 2023 (Online) 8th April Morning Shift
+4
-1
Out of Syllabus

If the equation of the plane containing the line

$$x+2 y+3 z-4=0=2 x+y-z+5$$ and perpendicular to the plane

$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$

is $a x+b y+c z=4$, then $$(a-b+c)$$ is equal to :

A
18
B
22
C
20
D
24
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