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Q.

The P.V.'s of the vertices of a triangle 2i+3j+4k, 4i+6j+3k, 3i+2j+3k, The P.V. of the orthocentre is

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a

2i-3j+4k

b

2i+3j-4k

c

2i+3j+4k

d

-2i+3j+4k

answer is C.

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Detailed Solution

OA=2i+3j+4k     OB=4i+4 OC=3i+2j+3k AB=2i-3j-k AB=14 BC =-i-4j    BC=1+4j=14 CA=-i+j+l  CA=1+j+l=3 ABC is right angle at A is orthocentre = OA

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