The plane 4x + 7y + 4z + 81 = 0 is rotated th | vedclass

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(A) The equation of any plane passing through the line of intersection of the planes $P_1: 4x + 7y + 4z + 81 = 0$ and $P_2: 5x + 3y + 10z - 25 = 0$ is

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$106$

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(A) The equation of any plane passing through the line of intersection of the planes $P_1: 4x + 7y + 4z + 81 = 0$ and $P_2: 5x + 3y + 10z - 25 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(4x + 7y + 4z + 81) + \lambda(5x + 3y + 10z - 25) = 0$$(4 + 5\lambda)x + (7 + 3\lambda)y + (4 + 10\lambda)z + (81 - 25\lambda) = 0$Since the new plane is $x - 4y + 6z - k = 0$,the normal vectors must be proportional:$\frac{4 + 5\lambda}{1} = \frac{7 + 3\lambda}{-4} = \frac{4 + 10\lambda}{6} = \frac{81 - 25\lambda}{-k}$From the first two ratios: $-16 - 20\lambda = 7 + 3\lambda \implies 23\lambda = -23 \implies \lambda = -1$.Since the planes are rotated by a right angle,the dot product of their normals must be zero:$n_1 \cdot n_2 = (4, 7, 4) \cdot (5, 3, 10) = 20 + 21 + 40 = 81 
eq 0$. Wait,the condition is that the new plane is perpendicular to the original plane $4x + 7y + 4z + 81 = 0$.$(4 + 5\lambda)(4) + (7 + 3\lambda)(7) + (4 + 10\lambda)(4) = 0$$16 + 20\lambda + 49 + 21\lambda + 16 + 40\lambda = 0 \implies 81\lambda + 81 = 0 \implies \lambda = -1$.Substituting $\lambda = -1$ into the ratio $\frac{81 - 25\lambda}{-k} = \frac{4 + 5\lambda}{1}$:$\frac{81 - 25(-1)}{-k} = \frac{4 + 5(-1)}{1} \implies \frac{106}{-k} = -1 \implies k = 106$.

The plane $4x + 7y + 4z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $5x + 3y + 10z = 25$. The equation of the plane in its new position is $x - 4y + 6z = k$,where $k$ is:

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