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(C) The equation of any plane passing through the line of intersection of planes $\pi_1 = 0$ and $\pi_2 = 0$ is given by $\pi_1 + \lambda \pi_2 = 0$.

Vedclass · Accepted Answer

$6x + 2y - 4z - 15 = 0$

Vedclass · Answer

(C) The equation of any plane passing through the line of intersection of planes $\pi_1 = 0$ and $\pi_2 = 0$ is given by $\pi_1 + \lambda \pi_2 = 0$. Substituting the given planes: $(2x + 6y + 4z - 7) + \lambda(x - y - 2z - 2) = 0$ $(2 + \lambda)x + (6 - \lambda)y + (4 - 2\lambda)z - (7 + 2\lambda) = 0 \quad \dots(i)$ Since this plane is perpendicular to the plane $x + y + 2z - 5 = 0$, the dot product of their normal vectors must be zero. The normal vectors are $\vec{n_1} = (2 + \lambda, 6 - \lambda, 4 - 2\lambda)$ and $\vec{n_2} = (1, 1, 2)$. $(2 + \lambda)(1) + (6 - \lambda)(1) + (4 - 2\lambda)(2) = 0$ $2 + \lambda + 6 - \lambda + 8 - 4\lambda = 0$ $16 - 4\lambda = 0 \implies \lambda = 4$. Substituting $\lambda = 4$ into equation $(i)$: $(2 + 4)x + (6 - 4)y + (4 - 8)z - (7 + 8) = 0$ $6x + 2y - 4z - 15 = 0$.

The equation of the plane passing through the line of intersection of planes $\pi_1: 2x + 6y + 4z - 7 = 0$ and $\pi_2: x - y - 2z - 2 = 0$, and perpendicular to the plane $x + y + 2z - 5 = 0$ is:

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