Let two planes be P_1 : 2x - y + z = 2 and P_ | vedclass

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(D) The equations of the planes are $P_1 : 2x - y + z - 2 = 0$ and $P_2 : x + 2y - z - 3 = 0$.To find the angle bisector,we use the formula $\frac{a_1

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$3x + y = 5$

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(D) The equations of the planes are $P_1 : 2x - y + z - 2 = 0$ and $P_2 : x + 2y - z - 3 = 0$.To find the angle bisector,we use the formula $\frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \pm \frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}}$.Substituting the values: $\frac{2x - y + z - 2}{\sqrt{4 + 1 + 1}} = \pm \frac{x + 2y - z - 3}{\sqrt{1 + 4 + 1}}$.Since $\sqrt{6} = \sqrt{6}$,we have $2x - y + z - 2 = \pm (x + 2y - z - 3)$.Case $1$ (Positive sign): $2x - y + z - 2 = x + 2y - z - 3 \Rightarrow x - 3y + 2z + 1 = 0$.Case $2$ (Negative sign): $2x - y + z - 2 = -x - 2y + z + 3 \Rightarrow 3x + y - 5 = 0$.To check which plane contains the origin $(0,0,0)$,we substitute $(0,0,0)$ into the equations:For $x - 3y + 2z + 1 = 0$,$0 - 0 + 0 + 1 = 1 
eq 0$.For $3x + y - 5 = 0$,$0 + 0 - 5 = -5 
eq 0$.Since both do not contain the origin,we look for the bisector of the angle containing the origin. The sign of $d_1$ and $d_2$ should be the same. Here $d_1 = -2$ and $d_2 = -3$. Since they have the same sign,the positive sign gives the bisector of the angle containing the origin. Thus,the negative sign gives the bisector not containing the origin: $3x + y - 5 = 0$.

Let two planes be $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Find the equation of the angle bisector plane of $P_1$ and $P_2$ that does not contain the origin.

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