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(B) The equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by:$(x + 2y + 3z

Vedclass · Accepted Answer

$51x + 15y - 50z + 173 = 0$

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(B) The equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by:$(x + 2y + 3z - 4) + \lambda (2x + y - z + 5) = 0$Rearranging the terms,we get:$x(1 + 2\lambda) + y(2 + \lambda) + z(3 - \lambda) + (5\lambda - 4) = 0 \quad \dots(i)$Since this plane is perpendicular to the plane $5x + 3y + 6z + 8 = 0$,the dot product of their normal vectors must be zero:$5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0$$5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$$7\lambda + 29 = 0$$\lambda = -\frac{29}{7}$Substituting $\lambda = -\frac{29}{7}$ into equation $(i)$:$(x + 2y + 3z - 4) - \frac{29}{7}(2x + y - z + 5) = 0$$7(x + 2y + 3z - 4) - 29(2x + y - z + 5) = 0$$7x + 14y + 21z - 28 - 58x - 29y + 29z - 145 = 0$$-51x - 15y + 50z - 173 = 0$$51x + 15y - 50z + 173 = 0$

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

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