The point of intersection of the lines `(x-5)/3=(y-7)/(-1)=(z+2)/1a n d=(x+3)/(-36)=(y-3)/2=(z-6)/4` is

Options

A. ` ( 2 , 10 , - 4 ) `
B. ` (21, ( 5 ) / ( 3 ) , ( 10 ) /( 3 )) `
C. ` ( 5, 7 , - 2 ) `
D. ` ( - 3, 3 , 6 ) `

Answer

B

Explanation

Given equation of lines are
` ( x - 5 ) / ( 3 ) = ( y - 7 ) / ( - 1 ) = ( z + 2 ) / ( 1 ) = k " "[say]" "`…(i)
` and ( x + 3 ) / ( - 36 ) = ( y - 3 ) / ( 2 ) = ( z - 6 ) / ( 4) " " `…(ii)
Any point on the line (i) is
` P ( 3k + 5, - k +7 , k - 2 ) `
This point is satisfied the Eq. (ii),
` therefore ( 3k + 5 + 2 ) / ( - 36 ) = ( - k + 7 - 3 ) / ( 2 ) = ( k - 2 - 6 ) / ( 4 ) `
` rArr ( 3k + 8 ) /( - 36 ) = ( - k + 4 )/ ( 2 ) = ( k- 8 ) / ( 4 )`
` rArr 3k + 8 = 18 k - 72 rArr k = ( 16) / ( 3 ) `
` therefore P ( 16 + 5, - ( 16 ) / (3 ) + 7, ( 16 ) / ( 3 ) - 2 ) `
i.e., ` P ( 21, ( 5) / ( 3 ) , ( 10 ) / ( 3 ) ) `