Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane passing through three non-collinear points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, and $(x_3,y_3,z_3)$ is given by the determinant:

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

Substituting the given points $(3,1,1)$, $(1,2,3)$, and $(-1,4,2)$:

$\left|\begin{array}{ccc}x-3& y-1& z-1\\ 1-3& 2-1& 3-1\\ -1-3& 4-1& 2-1\end{array}\right|=0$

$\left|\begin{array}{ccc}x-3& y-1& z-1\\ -2& 1& 2\\ -4& 3& 1\end{array}\right|=0$

Expanding the determinant along the first row:

$(x-3)[(1)(1)-(2)(3)]-(y-1)[(-2)(1)-(2)(-4)]+(z-1)[(-2)(3)-(1)(-4)]=0$

$(x-3)(1-6)-(y-1)(-2+8)+(z-1)(-6+4)=0$

$-5(x-3)-6(y-1)-2(z-1)=0$

$-5x+15-6y+6-2z+2=0$

$-5x-6y-2z+23=0$

Multiplying by $-1$, we get the equation of the plane as:

$5x+6y+2z-23=0$

Answer: $5x+6y+2z-23=0$