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Question

Find the coordinates of those points on the line x+12=y+23=z36 which is at a distance of 3 units from the point (1,2,3)

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Solution

x+12=y+23=z36
Let the point be
=(2k1,3k2,6k+3)
Distance from (1,2,3)
=(2k11)2+(3k2+2)2+(6k+33)2
=(2k22+9k2+36k2=3
4k2+48k+45k2=9
49k28k5=0
k=8±64+20×4998
k=8±65+98098
k=8±3298=4098=2049
Point are (949,3849,26749)

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