The distance between the plane whose equation is →r⋅(2→i+→j−3→k)=5 and the line whose equation is →r=→i+λ(2→i+5→j+3→k),is
A
3√14
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B
5√14
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C
5
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D
0
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Solution
The correct option is A3√14 Clearly line and plane are parallel, so perpendicular distance between line and plane will be same at each point Now equation of line in cartesian form is 2x+y−3z−5=0 and the line passes through (1,0,0) So distance between the line and the plane is =∣∣∣2(1)−522+12+32∣∣∣=3√14