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

Vedclass · Accepted Answer

$x + 3y + 6z = 7$

Vedclass · Answer

(D) The equation of the family of planes passing through the intersection of the planes $2x - 5y + z - 3 = 0$ and $x + y + 4z - 5 = 0$ is given by:$(2x - 5y + z - 3) + \lambda(x + y + 4z - 5) = 0$Rearranging the terms,we get:$(2 + \lambda)x + (\lambda - 5)y + (4\lambda + 1)z - (3 + 5\lambda) = 0 \quad \dots(i)$Since this plane is parallel to the plane $x + 3y + 6z = 1$,the normal vectors must be proportional:$\frac{2 + \lambda}{1} = \frac{\lambda - 5}{3} = \frac{4\lambda + 1}{6} = k$From $\frac{2 + \lambda}{1} = \frac{\lambda - 5}{3}$,we have $6 + 3\lambda = \lambda - 5 \implies 2\lambda = -11 \implies \lambda = -\frac{11}{2}$.Substituting $\lambda = -\frac{11}{2}$ into equation $(i)$:$(2 - \frac{11}{2})x + (-\frac{11}{2} - 5)y + (4(-\frac{11}{2}) + 1)z - (3 + 5(-\frac{11}{2})) = 0$$-\frac{7}{2}x - \frac{21}{2}y - 21z - (3 - \frac{55}{2}) = 0$$-\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0$Multiplying by $-\frac{2}{7}$,we get:$x + 3y + 6z - 7 = 0 \implies x + 3y + 6z = 7$.

The equation of the plane containing the line $2x - 5y + z = 3; x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is:

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