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The symmetric equation of lines 3x+2z-5=0 and x+y-2z-3=0 isA. `(x-1)/(5)=(y-4)/(7)=(z-0)/(1)`B. `(x+1)/(5)=(y+4)/(7)=(z-0)/(1)`C. `(x+1)/(-5)=(y+4)/(7)=(z-0)/(1)`D. `(x-1)/(-5)=(y-4)/(7)=(z-0)/(1)`

Answer» Correct Answer - C
Let a,b,c be the direction ratios of required line.
`therefore 3a+2b+c=0`
and a+b-2c=0
`Rightarrow (a)/(-4-1)=(b)/(1+6)=(c)/(3-2)`
`Rightarrow (a)/(-5)=(b)/(7)=(c)/(1)`
In order to find on the required line, we put z=0 in the two given equations to obtain 3x+2y=5 and x+y=3.
On solving these two equations, we get x=-1, y=4
`therefore` Coordintaes of point on required line are (-1,4,0)
Hence, required line is `(x+1)/(-5)=(y-4)/(7)=(z-0)/(1)`


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