Question

Find the equation of the plane passing
through (3, 5, 1), (2, 3, 0), (0, 6, 0).​

Answer 1

Answer:

$\bold\red{3x+2y-7z-12=0}$

Step-by-step explanation:

The Equation of plane passing through

(3,5,1) , (2,3,0) and (0,6,0) is given by

$(x - 3)(2 + 1) - (y - 5)(1 - 3) + (z - 1)( - 1 - 6) = 0 \\ \\ = > 3(x - 3) + 2(y - 5) - 7(z - 1) = 0 \\ \\ = > 3x + 2y - 7z + (7 - 10 - 9) = 0 \\ \\ = > 3x + 2y - 7z - 12 = 0$

Hence,

the required equation of plane is

$\bold{3x+2y-7z-12=0}$

Answer 2

Answer:

Step-by-step explanation:

The Equation of plane passing through

(3,5,1) , (2,3,0) and (0,6,0) is given by

(x - 3)(2   + 1) - (y - 5)(1 - 3) + (z - 1)( - 1 - 6) = 0 \\  \\  =  > 3(x - 3) + 2(y - 5)  - 7(z  - 1) = 0 \\  \\  =  > 3x + 2y - 7z + (7 - 10 - 9) = 0 \\  \\  =  > 3x + 2y - 7z - 12 = 0

Hence,

the required equation of plane is

\bold{3x+2y-7z-12=0}

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