The distance of the point (1, 3,-7) from the plane passing through the point (1,-1,-1) , having normal perpendicular to both the lines x-1 1= y+2 -2=…
JEE Main 2017 — Mathematics Vectors & 3D Geometry
2017mcqmedium
The distance of the point (1,3,−7) from the plane passing through the point (1,−1,−1) , having normal perpendicular to both the lines 1x−1=−2y+2=3z−4 and 2x−2=−1y+1=−1z+7 , is:
Official previous-year question
Held on 2 Apr 2017 · Verified 6 Jul 2026.
Options
A
7420
B
8310
C
835
D
7410
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Solution
Let l,m,n be the direction cosines of the line normal to the plane, then
l−2m+3n=0 and 2l−m−n=0
⇒2+3l=6+1m=−1+4n=λ
⇒=5λ,m=7λ,n=3λ
∴ The equation of the plane is
5x+7y+3y+d=0
∵it passes through(1,−1,−1)
⇒5−7−3+d=0
⇒d=5
Hence, the equation of the plane is 5x+7y+3y+5=0
Now, ∣PQ∣=(5)2+(7)2+(3)2∣5+21−21+5∣
⇒∣PQ∣=25+49+910
⇒∣PQ∣=8310
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