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If the lines `(x-1)/2=(y+1)/3=(z-1)/4` and `(x-3)/1=(y-k)/2=z/1` intersect, then k is equal to (1) `-1` (2) `2/9` (3) `9/2` (4) 0
A. ` - 1 `
B. ` ( 2 ) / ( 9 ) `
C. ` ( 9 ) / ( 2 ) `
D. ` 0 `

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Correct Answer - C
` L _ 1 : ( x - 1 ) / ( 2 ) = ( y + 1 ) / ( 3 ) = ( z - 1 ) / ( 4 ) = p `
` L _ 2 : ( x - 3 ) / ( 1) = ( y - k) /( 2 ) = ( z - 0 ) / ( 1 ) = q `
` rArr ` Anypoint P on line ` L _ 1 ` is of type ` P ( 2p + 1, 3p -1, 4p + 1 ) ` and any point Q on line ` L _ 2 ` is type Q ` ( q + 3, 2q + k , q ) `
Since, ` L _ 1 and L _ 2 ` are intersecting each other, hence both point P and Q should coincide at the point of intersection i.e., corresponding coordinates of P and Q should be same.
` 2p + 1 = q + 3 and 4 p + 1 = q `, we get the value of p and q as
` p = ( - 3 ) / ( 2 ) and q = - 5 `
On substituting the values of p and q in the third equation ` 3p - 1 = 2q + k `, we get
` therefore 3 ( ( - 3 ) / ( 2 ) ) - 1 = 2 ( - 5) + k`
` rArr k = ( 9 ) / ( 2 ) `

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