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(D) The equations of the planes are $P_1: 3x - 6y + 2z + 5 = 0$ and $P_2: 4x - 12y + 3z - 3 = 0$. To find the bisector containing the origin,we first

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None of these

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(D) The equations of the planes are $P_1: 3x - 6y + 2z + 5 = 0$ and $P_2: 4x - 12y + 3z - 3 = 0$. To find the bisector containing the origin,we first make the constant terms positive. For $P_1$,the constant is $5 > 0$,so we keep it as $3x - 6y + 2z + 5 = 0$. For $P_2$,the constant is $-3 The equation of the bisector containing the origin is given by $\frac{3x - 6y + 2z + 5}{\sqrt{3^2 + (-6)^2 + 2^2}} = \frac{-4x + 12y - 3z + 3}{\sqrt{(-4)^2 + 12^2 + (-3)^2}}$. This simplifies to $\frac{3x - 6y + 2z + 5}{7} = \frac{-4x + 12y - 3z + 3}{13}$. Cross-multiplying gives $13(3x - 6y + 2z + 5) = 7(-4x + 12y - 3z + 3)$. $39x - 78y + 26z + 65 = -28x + 84y - 21z + 21$. Rearranging terms: $67x - 162y + 47z + 44 = 0$. Since this result is not among the options,the correct choice is $D$.

The equation of the plane which bisects the angle between the planes $3x - 6y + 2z + 5 = 0$ and $4x - 12y + 3z - 3 = 0$ which contains the origin is

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