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(C) Let the equation of the plane passing through the origin be $ax+by+cz=0$. Since this plane is perpendicular to the planes $x+2y-z=1$ and $3x-4y+z=

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$x+2y+5z=0$

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(C) Let the equation of the plane passing through the origin be $ax+by+cz=0$. Since this plane is perpendicular to the planes $x+2y-z=1$ and $3x-4y+z=5$,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal vectors of the given planes,$\vec{n_1} = (1, 2, -1)$ and $\vec{n_2} = (3, -4, 1)$. Thus,$\vec{n} = \vec{n_1} 	imes \vec{n_2}$. $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 3 & -4 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1+3) + \hat{k}(-4-6) = -2\hat{i} - 4\hat{j} - 10\hat{k}$. We can take the normal vector as $\vec{n} = (1, 2, 5)$ by dividing by $-2$. The equation of the plane is $1(x-0) + 2(y-0) + 5(z-0) = 0$,which simplifies to $x+2y+5z=0$.

The equation of the plane passing through the origin and perpendicular to the planes $x+2y-z=1$ and $3x-4y+z=5$ is:

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