The distance of the point P(3,4,4) from the point of intersection of the line joining the points Q(3,-4,-5) and R(2,-3,1) and the plane 2x+y+z=7, is…
JEE Main 2021 — Mathematics Vectors & 3D Geometry
2021integermedium
The distance of the point P(3,4,4) from the point of intersection of the line joining the points Q(3,−4,−5) and R(2,−3,1) and the plane 2x+y+z=7, is equal to _____.
Official previous-year question
Held on 27 Jul 2021 · Verified 6 Jul 2026.
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Solution
We have, point P(3,4,4)
Equation of line joining the points Q(3,−4,−5) and R(2,−3,1) is −1x−3=1y+4=6z+5=r
Any point on above line is (−r+3,r−4,6r−5)
Now, satisfying it in the given plane 2x+y+z=7, we get
2(−r+3)+(r−4)+(6r−5)=7
⇒r=2
So, required point of intersection is T(1,−2,7).
Hence, PT=(3−1)2+(4+2)2+(4−7)2=7.
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