Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).


We are given with points, (2, 1, 0), (3, -2, -2) and (3, 1, 7).

We know that, the equation of a line passing through three non-collinear points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given as



Here,


(x1, y1, z1) ≡ (2, 1, 0)


(x2, y2, z2) ≡ (3, -2, -2)


(x3, y3, z3) ≡ (3, 1, 7)


So,


x1 = 2, y1 = 1 and z1 = 0


x2 = 3, y2 = -2 and z2 = -2


x3 = 3, y3 = 1 and z3 = 7


Putting these values in the equation of the line, we get




Take 1st row and 1st column, multiply the first element of the row (a11) with the difference of multiplication of opposite elements (a22 × a33 – a23 × a32), excluding 1st row and 1st column.



Here,



Now take 1st row and 2nd column, multiply the second element of the row (a12) with the difference of multiplication of opposite elements (a21 × a33 – a23 × a31), excluding 1st row and 2nd column.



Here,



Similarly, take 1st row and 3rd column, multiply the third element of the row (a13) with the difference of multiplication of opposite elements (a22 × a33 – a23 × a32), excluding 1st row and 3rd column.



Here,



Solving it further,








Now, since


-21x – 9y + 3z + 51 = 0


-21x – 9y + 3z = -51


-3(7x + 3y – z) = -3 × 17


7x + 3y – z = 17


Thus, required equation of the plane is 7x + 3y – z = 17.


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