Show that the plane through the points (1,1,1), (1,-1,1), (-7,3,-5) is perpendicular to the xz-plane

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$$\begin{gathered} \mathrm{Let} \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{be} \mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0 \\ \mathrm{We} \mathrm{have}: \\ \mathrm{A}\left(1,1,1\right), \mathrm{B}\left(1,-1,1\right) \mathrm{and} \mathrm{C}\left(-7,3,-5\right) \\ \overrightarrow{\mathrm{AB}}=\mathrm{B}-\mathrm{A}=\left(1-1\right)\hat{\mathrm{i}}+\left(-1-1\right)\hat{\mathrm{j}}+\left(1-1\right)\hat{\mathrm{k}}=-2\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{AC}}=\mathrm{C}-\mathrm{A}=\left(-7-1\right)\hat{\mathrm{i}}+\left(3-1\right)\hat{\mathrm{j}}+\left(-5-1\right)\hat{\mathrm{k}}=-8\hat{\mathrm{i}}+2\hat{\mathrm{j}}-6\hat{\mathrm{k}} \\ \mathrm{Cross} \mathrm{product} \mathrm{of} \mathrm{above} \mathrm{vectors} \mathrm{will} \mathrm{give} \mathrm{direction} \mathrm{normals} \\ \overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 0& -2& 0\\ -8& 2& -6\end{array}\right| \\ =\hat{\mathrm{i}}\left(12-0\right)-\hat{\mathrm{j}}\left(0-0\right)+\hat{\mathrm{k}}\left(0-16\right) \\ =12\hat{\mathrm{i}}+0\hat{\mathrm{j}}-16\hat{\mathrm{k}} \\ \mathrm{As} \mathrm{coefficient} \mathrm{of} \hat{\mathrm{j}} \mathrm{is} 0, \mathrm{hence} \mathrm{plane} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{xz} \mathrm{plane}. \end{gathered}$$

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