Find the equation of the plane passing through (2,3,1) and perpendicular to the line joining the points (3,4,1) and (2,1,5) ?
I'm assuming the points are (3.4,1), (2,1.5) and (2.3,1)
Using point gradient form,
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To find the equation of the plane passing through the point (2,3,1) and perpendicular to the line joining the points (3,4,1) and (2,1,5), follow these steps:
 Find the direction vector of the line joining the two points.
 Use the point (2,3,1) and the direction vector to form the normal vector of the plane.
 Use the pointnormal form of the equation of a plane to find the equation.
Let's calculate:

Direction vector of the line joining the points (3,4,1) and (2,1,5): [ \text{Direction vector} = (23, 14, 5+1) = (1, 5, 6) ]

Using the point (2,3,1) and the direction vector (1, 5, 6), we find the normal vector of the plane: [ \text{Normal vector} = (1, 5, 6) ]

The equation of the plane in pointnormal form is given by: [ A(x  x_0) + B(y  y_0) + C(z  z_0) = 0 ] where ( (x_0, y_0, z_0) ) is the given point (2,3,1).
Substitute the values: [ 1(x  2)  5(y  3) + 6(z + 1) = 0 ]
Expanding and simplifying: [ x + 2 + 5y  15 + 6z + 6 = 0 ] [ x + 5y + 6z  7 = 0 ]
So, the equation of the plane passing through the point (2,3,1) and perpendicular to the line joining the points (3,4,1) and (2,1,5) is ( x + 5y + 6z  7 = 0 ).
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