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(A) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.Here,$(x + y +

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$20x + 23y + 26z - 69 = 0$

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(A) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.Here,$(x + y + z - 6) + \lambda (2x + 3y + 4z + 5) = 0$.Since the plane passes through the point $(1, 1, 1)$,we substitute $x=1, y=1, z=1$ into the equation:$(1 + 1 + 1 - 6) + \lambda (2(1) + 3(1) + 4(1) + 5) = 0$$(-3) + \lambda (2 + 3 + 4 + 5) = 0$$-3 + 14\lambda = 0$$14\lambda = 3 \Rightarrow \lambda = \frac{3}{14}$.Substituting $\lambda = \frac{3}{14}$ back into the equation:$(x + y + z - 6) + \frac{3}{14} (2x + 3y + 4z + 5) = 0$$14(x + y + z - 6) + 3(2x + 3y + 4z + 5) = 0$$14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0$$20x + 23y + 26z - 69 = 0$.

Find the equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$.

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