1
JEE Main 2019 (Online) 9th April Evening Slot
+4
-1
The vertices B and C of a $$\Delta$$ABC lie on the line,

$${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$$ such that BC = 5 units.

Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :
A
6
B
$$5\sqrt {17}$$
C
$$\sqrt {34}$$
D
$$2\sqrt {34}$$
2
JEE Main 2019 (Online) 9th April Evening Slot
+4
-1
Out of Syllabus
Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :
A
205$$\sqrt5$$
B
63$$\sqrt5$$
C
11/$$\sqrt5$$
D
17/$$\sqrt5$$
3
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
Out of Syllabus
A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y – z + 5 = 0, also passes through the point
A
$$\left( {\sqrt 2 ,1,4} \right)$$
B
$$\left(- {\sqrt 2 ,1,4} \right)$$
C
$$\left( -{\sqrt 2 ,-1,-4} \right)$$
D
$$\left( {\sqrt 2 ,-1,4} \right)$$
4
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
Out of Syllabus
If the line, $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$$ meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from the origin is :
A
$${{\sqrt 5 } \over 2}$$
B
2$$\sqrt 5$$
C
9/2
D
7/2
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