The equation of a plane containing the line of intersection of the planes `2x-y-4=0` and `y+2z-4=0` and passing through the point (1, 1, 0) isA. `x-3y-2z=-2`B. `2x-z=2`C. `x-y-z=0`D. `x+3y+z=4`

The equation of a plane containing the line of intersection of the pla

1.	The equation of a plane containing the line of intersection of the planes `2x-y-4=0` and `y+2z-4=0` and passing through the point (1, 1, 0) isA. `x-3y-2z=-2`B. `2x-z=2`C. `x-y-z=0`D. `x+3y+z=4`
Answer» Correct Answer - C Equations of given planes are `2x-y-4=0" "…(i)` and`" "y+2z-4=0" "…(ii)` Now, equation of family of planes passes through the line of intersection of given planes (i) and (ii) is `(2x-y-4)+lambda(y+2z-4)=0" "...(iii)` According to the question, Plane (iii) passes through the point, (1, 1, 0), so `(2-1-4)+lambda(1+0-4)=0` `implies" "-3-3lambda=0` `implies" "lambda=-1` Now, equation of required plane can be obtained by putting`lambda=-1` in the equation of plane (iii). `implies(2x-y-4)-1(y+2z-4)=0` `implies" "2x-y-4-y-2z+4=0` `implies" "2x-2y-2z=0` `impliesx-y-z=0`

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The equation of a plane containing the line of intersection of the planes `2x-y-4=0` and `y+2z-4=0` and passing through the point (1, 1, 0) isA. `x-3y-2z=-2`B. `2x-z=2`C. `x-y-z=0`D. `x+3y+z=4`

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