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Question

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

Vedclass · Accepted Answer

$7x - 5y + 4z - 8 = 0$

Vedclass · Answer

(A) The equation of any plane passing through the intersection of the planes $P_1: 3x - y + 2z - 4 = 0$ and $P_2: x + y + z - 2 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(3x - y + 2z - 4) + \lambda(x + y + z - 2) = 0$  --- $(1)$Since the plane passes through the point $(2, 2, 1)$,this point must satisfy the equation $(1)$.Substituting $x = 2, y = 2, z = 1$ in equation $(1)$:$(3(2) - 2 + 2(1) - 4) + \lambda(2 + 2 + 1 - 2) = 0$$(6 - 2 + 2 - 4) + \lambda(3) = 0$$2 + 3\lambda = 0$$\lambda = -\frac{2}{3}$Substituting $\lambda = -\frac{2}{3}$ back into equation $(1)$:$(3x - y + 2z - 4) - \frac{2}{3}(x + y + z - 2) = 0$$3(3x - y + 2z - 4) - 2(x + y + z - 2) = 0$$(9x - 3y + 6z - 12) - (2x + 2y + 2z - 4) = 0$$7x - 5y + 4z - 8 = 0$Thus,the required equation of the plane is $7x - 5y + 4z - 8 = 0$.

Find the equation of the plane passing through the intersection of the planes $3x - y + 2z - 4 = 0$ and $x + y + z - 2 = 0$ and the point $(2, 2, 1)$.

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