Find the equation of the plane through the points `(23,1)a n d(4,-5,3)` and parallel to the x-axis.

Explanation

Plane passes through the points `(2, 3, 1) and (4, -5, 3)`
Therefore, plane is parallel to vector
`" "(4-2)hati+(-5-3)hatj+ (3-1)hatk=2hati-8hatj+2hatk`
Also plane is parallel to the x-axis, so it parallel to vector `hati`
`therefore` Normal to the plane = `|{:(hati,,hatj,,hatk),(2,,-8,,2),(1,,0,,0):}|=2hatj+8hatk`
Hence, equation of plane through point (2, 3, 1) and normal to vector `2hatj+8hatk` is
`" "0(x-2)+2(y-3)+8(z-1)=0`
or `" "y+4z-7=0`