The equation of the line of intersection of the planes `x+2y+z=3` and `6x+8y+3z=13` can be written as

Options

A. `(x-2)/(2)=(y+1)/(-3)=(z-3)/(4)`
B. `(x-2)/(2)=(y+1)/(3)=(z-3)/(4)`
C. `(x+2)/(2)=(y-1)/(-3)=(z-3)/(4)`
D. `(x+2)/(2)=(y+2)/(3)=(z-3)/(4)`

Answer

A

Explanation

Let the Dr's of a required line a,b and c. Since, the normal to the given planes `x+2y+z=3and 6x+8y+3z=13` are perpendicular to the line
`:.a+2b+c=0" "[becausea_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0]`
`and" "6a+8b+3c=0`
`rArr" "(a)/(6-8)=(b)/(6-3)=(c)/(8-12)`
`rArr" "(a)/(-2)=(b)/(3)=(c)/(-4)`
`or" "(a)/(2)=(b)/(-3)=(c)/(4)`
Also, line passes through (2,-1,3).
`:.` Equation is `(x-2)/(2)=(y+1)/(-3)=(z-3)/(4)`.