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MATHEMATICS · Single correct · CBSE · Mathematics · Three-Dimensional Geometry · Three-Dimensional Geometry
Problem The distance of the point ( 1 , 0 , 2 ) (1, 0, 2) ( 1 , 0 , 2 ) x − 2 3 = y + 1 4 = z − 2 12 \frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} 3 x − 2 = 4 y + 1 = 12 z − 2 x − y + z = 16 x - y + z = 16 x − y + z = 16 2 14 2\sqrt{14} 2 14 3 21 3\sqrt{21} 3 21 Browse the problem library