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(C) The equation of any plane passing through the intersection of the planes $P_1: 2x - 5y + z - 3 = 0$ and $P_2: x + y + 4z - 5 = 0$ is given by $P_1

Vedclass · Accepted Answer

$7$

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(C) The equation of any plane passing through the intersection of the planes $P_1: 2x - 5y + z - 3 = 0$ and $P_2: x + y + 4z - 5 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(2x - 5y + z - 3) + \lambda(x + y + 4z - 5) = 0$Rearranging the terms,we get:$(2 + \lambda)x + (-5 + \lambda)y + (1 + 4\lambda)z - (3 + 5\lambda) = 0 \dots (1)$Since this plane is parallel to the plane $x + 3y + 6z - 1 = 0$,the normal vectors must be proportional:$\frac{2 + \lambda}{1} = \frac{-5 + \lambda}{3} = \frac{1 + 4\lambda}{6}$Taking the first two parts:$3(2 + \lambda) = -5 + \lambda$$6 + 3\lambda = -5 + \lambda$$2\lambda = -11 \Rightarrow \lambda = -\frac{11}{2}$Substituting $\lambda = -\frac{11}{2}$ into equation $(1)$:$(2 - \frac{11}{2})x + (-5 - \frac{11}{2})y + (1 - 22)z - (3 - \frac{55}{2}) = 0$$-\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0$Multiplying by $-2$:$7x + 21y + 42z - 49 = 0$$7(x + 3y + 6z) = 49$$x + 3y + 6z = 7$Comparing this with $x + 3y + 6z = k$,we get $k = 7$.

The equation of the plane passing through the intersection of the planes $2x - 5y + z = 3$ and $x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is $x + 3y + 6z = k$,where $k$ is:

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