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

Vedclass · Accepted Answer

$5x - 11y + z = 17$

Vedclass · Answer

(A) The equation of any plane passing through the intersection of the planes $P_1: x + 2y + 3z - 2 = 0$ and $P_2: x - y + z - 3 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(x + 2y + 3z - 2) + \lambda(x - y + z - 3) = 0$$(1 + \lambda)x + (2 - \lambda)y + (3 + \lambda)z - (2 + 3\lambda) = 0$The distance of this plane from the point $(3, 1, -1)$ is given as $\frac{2}{\sqrt{3}}$.Using the distance formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$,we have:$\frac{|(1 + \lambda)(3) + (2 - \lambda)(1) + (3 + \lambda)(-1) - (2 + 3\lambda)|}{\sqrt{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2}} = \frac{2}{\sqrt{3}}$$|3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda| = | -2\lambda |$Denominator: $\sqrt{1 + 2\lambda + \lambda^2 + 4 - 4\lambda + \lambda^2 + 9 + 6\lambda + \lambda^2} = \sqrt{3\lambda^2 + 4\lambda + 14}$So,$\frac{|-2\lambda|}{\sqrt{3\lambda^2 + 4\lambda + 14}} = \frac{2}{\sqrt{3}}$Squaring both sides: $\frac{4\lambda^2}{3\lambda^2 + 4\lambda + 14} = \frac{4}{3}$$3\lambda^2 = 3\lambda^2 + 4\lambda + 14$$4\lambda = -14 \Rightarrow \lambda = -\frac{7}{2}$Substituting $\lambda = -\frac{7}{2}$ into the plane equation:$(1 - \frac{7}{2})x + (2 + \frac{7}{2})y + (3 - \frac{7}{2})z - (2 - \frac{21}{2}) = 0$$-\frac{5}{2}x + \frac{11}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$$-5x + 11y - z + 17 = 0 \Rightarrow 5x - 11y + z = 17$.

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

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