The plane passing through the points (1,1,1), | vedclass

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(B) The equation of the plane passing through the 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 equatio

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parallel to $Y$-axis

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(B) The equation of the plane passing through the 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 equation: $\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}ight| = 0$ Substituting the given points $(1,1,1)$,$(1,-1,1)$ and $(-7,-3,-5)$: $\left|\begin{array}{ccc} x-1 & y-1 & z-1 \ 1-1 & -1-1 & 1-1 \ -7-1 & -3-1 & -5-1 \end{array}ight| = 0$ $\left|\begin{array}{ccc} x-1 & y-1 & z-1 \ 0 & -2 & 0 \ -8 & -4 & -6 \end{array}ight| = 0$ Expanding along the first row: $(x-1)((-2)(-6) - (0)(-4)) - (y-1)((0)(-6) - (0)(-8)) + (z-1)((0)(-4) - (-2)(-8)) = 0$ $(x-1)(12) - (y-1)(0) + (z-1)(-16) = 0$ $12x - 12 - 16z + 16 = 0$ $12x - 16z + 4 = 0$ Dividing by $4$: $3x - 4z + 1 = 0$ Since the $y$-coefficient is $0$,the normal vector is $(3, 0, -4)$,which is perpendicular to the $Y$-axis. Therefore,the plane is parallel to the $Y$-axis.

The plane passing through the points $(1,1,1)$,$(1,-1,1)$ and $(-7,-3,-5)$ is

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