Question

Find the equation of the plane through (-1,6,2) and perpendicular to the join of
(1,2,3) and (-2,3,4).​

Answer 1

The equation of plane is -3x + y + z = 11 .

We need to Find the equation of the plane through (-1,6,2) and perpendicular to the joining of   (1,2,3) and (-2,3,4) .​

We know equation of line passing through a point  $(x_1,y_1,z_1)$ and perpendicular to line passing through points $(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is :

$(x_3-x_2)(x-x_1)+(y_3-y_2)(y-y_1)+(z_3-z_2)(z-_1)=0$

Putting all those value we get .

$(-2-1)(x+1)+(3-2)(y-6)+(4-3)(z-2)=0\\\\-3x-3+y-6+z-2=0\\\\-3x+y+z=11$

Hence , this is the required solution .

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