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Question

The equation of line AB is x2=y−3=z6. Through a point P(1,2,5), line PN is drawn perpendicular to AB and line PQ is drawn parallel to the plane 3x+4x+5z=0 to meet AB at Q. Then

A
Coordinates of N are (5249,7849,15649)
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B
Equation of line NQ is 3(x3)=(2y+9)=z9
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C
Equation of line NQ is 3(x3)=(2y+9)=z9
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D
Coordinates of Q are (3,92,9)
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Solution

The correct options are
A Coordinates of N are (5249,7849,15649)
B Equation of line NQ is 3(x3)=(2y+9)=z9
Since, N lies on line AB
and direction ratios of line AB is <2,3,6>
So, coordinates of N(2r,3r,6r)
(where rR be any constant)
d.r.'s of PN is <2r1,3r2,6r5>
Since, PNAB
2(2r1)3(3r2)+6(6r5)=0r=2649
So, N(5249,7849,15649)

Now, let coordinates of Q be (2k,3k,6k)
(where kR be any constant)
and d.r.'s of perpendicular of the given plane is <3,4,5>
Then d.r.'s of PQ is <2k1,3k2,6k5>
Since, PQ plane
3(2k1)+4(3r2)+5(6k5)=0k=32
So, Q(3,92,9)
Equation of NQ is x395=y+9/2285/2=z9285

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