The plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4=0` and parallel to Y-axis also passes through the pointA. (3, 3, -1)B. (-3, 1, 1)C. (3, 2, 1)D. (-3, 0, -1)

The plane through the intersection of the planes `x+y+z=1` and `2x+3y-

1.	The plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4=0` and parallel to Y-axis also passes through the pointA. (3, 3, -1)B. (-3, 1, 1)C. (3, 2, 1)D. (-3, 0, -1)
Answer» Correct Answer - C Equation of poane through the intersection of two planes `P_(1)` and `P_(2)` is given by `P_(1)+lambdaP_(2)=0` The plane through the intersection of the planes `x+y+z-1=0` and `2x+3y-z+4=0` is given by `(x+y+z-1)+lambda(2x+3y-z+4)=0,` where `lambdainR` `implies(1+2lambda)x+(1+3lambda)y+(1-lambda)z+(4lambda-1)=0,` where `lambdainR" "...(i)` Since, this plane is parallel to Y-axis, therefore its normal is perpendicular to Y-axis, therefore its normal is perpendicular to Y-axis. `implies" "{(1+2lambda)hati+(1+3lambda)hatj+(1-lambda)hatk}.hatj=0` `implies" "1+3lambda=0implies" "lambda=-(1)/(3)` Now, required equation of plane is `(1-(2)/(3))x+(1-(3)/(3))y+(1+(1)/(3))z+(-(4)/(3)-1)=0` [substituting `lambda=(-1)/(3)` in Eq. (i)] `impliesx+4z-7=0` Here, only (3, 2, 1) satisfy the above equation.

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The plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4=0` and parallel to Y-axis also passes through the pointA. (3, 3, -1)B. (-3, 1, 1)C. (3, 2, 1)D. (-3, 0, -1)

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