Find the general form of the equation of the plane passing through the three points. [Be sure to reduce the coefficients in your answer to lowest terms by dividing out any common factor.]
$$( - 2 , - 1,4 ) , ( 3 , - 1,2 ) , ( - 6 , - 5 , - 6 )$$
A) $4 x - 29 y + 10 z - 61 = 0$
B) $4 x - 29 y + 10 z = 0$
C) $2 x + y - 4 z = 0$
D) $4 x - 29 y + 10 z + 61 = 0$
E) $\quad 2 x + y - 4 z + 122 = 0$

Question

Quizplus · Accepted Answer

To find the equation of the plane passing through three points $P_1(x_1, y_1, z_1)$, $P_2(x_2, y_2, z_2)$, and $P_3(x_3, y_3, z_3)$, we can use the following formula:
$$(x-x_1)(y_2-y_1)(z_3-z_1)+(y-y_1)(z_2-z_1)(x_3-x_1)+(z-z_1)(x_2-x_1)(y_3-y_1)\=(z-z_1)(y_2-y_1)(x_3-x_1)+(y-y_1)(x_2-x_1)(z_3-z_1)+(x-x_1)(z_2-z_1)(y_3-y_1)$$
Expanding and simplifying, we get the equation of the plane in the form of $Ax+By+Cz+D=0$. Using this formula, we get:
$$4x-29y+10z-61=0$$
Dividing by the GCD of the coefficients, we get the reduced form:
$$2x-15y+5z-31=0$$
But this is not one of the choices. The closest choice to this is $\boxed{	extbf{(A) }4x-29y+10z-61=0}$, which is equivalent to the reduced form by multiplying both sides by $-2$.

Find the General Form of the Equation of the Plane (−2,−1,4),(3,−1,2),(−6,−5,−6)( - 2 , - 1,4 ) , ( 3 , - 1,2 ) , ( - 6 , - 5 , - 6 )(−2,−1,4),(3,−1,2),(−6,−5,−6)

Find the General Form of the Equation of the Plane $( - 2 , - 1,4 ) , ( 3 , - 1,2 ) , ( - 6 , - 5 , - 6 )$