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Question

Let F be the foot of perpendicular from P(1,2,3) on the line L:x+12=y32=z1. Given Q(x1,y1,z1) and R(x2,y2,z2) are points at a distance of 3 units from F on line L. Let π1 and π2 be two planes passing through Q and R respectively and perpendicular to the line of intersection of the planes 9x7y+6z+48=0 and x+yz=7. Then which of the following statements is (are) CORRECT?

A
The coordinates of F are (1,1,1)
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B
The value of 2i=1(x2i+y2i+z2i) equals 24
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C
The distance between planes π1 and π2 equals 88482
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D
The normal vector of planes π1 and π2 is parallel to the vector ^i+15^j+16^k.
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Solution

The correct option is D The normal vector of planes π1 and π2 is parallel to the vector ^i+15^j+16^k.
L:x+12=y32=z1=λ (let)
x=2λ1,y=32λ,z=λ

Clearly, 2(2λ11)2(32λ2)1(λ+3)=0
2(2λ2)2(12λ)1(3λ)=0
4λ42+4λ3+λ=0 9λ=9λ=1
F(1,1,1)

Co-ordinates of a point on L at a distance r from (x1,y1,z1) is (x1+lr,y1+mr,z1+nr)
Here, l=23, m=23,n=13,r=±3 {a2+b2+c2=22+22+12=3}
So, Q(3,1,2)(x1,y1,z1) and R(1,3,0)(x2,y2,z2)
2i=1(x2i+y2i+z2i) =x21+y21+z21+x22+y22+z22 =(3)2+(1)2+(2)2+(1)2+(3)2+0
=24

Now, normal vector to planes π1 and π2 is n1×n2=n (let)
Given, n1=9i7j+6k,n2=i+jk
So, n=n1×n2=∣ ∣ijk976111∣ ∣
=^i(76)^j(96)+^k(9+7) =^i+15^j+16^k
Equation of planes π1 and π2 are 1(x3)+15(y+1)+16(z+2)=0 and 1(x+1)+15(y3)+16(z0)=0
i.e., x+15y+16z+44=0 and x+15y+16z44=0

Distance between planes π1 and π2 =∣ ∣44+4412+152+162∣ ∣=88482

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