The equation of a plane at a distance of sqrt | vedclass

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(D) The equation of any plane passing through the intersection of the planes $P_1: x-y-z-1=0$ and $P_2: 2x+y-3z+4=0$ is given by $P_1 + \lambda P_2 =

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$4x-y-5z+2=0$

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(D) The equation of any plane passing through the intersection of the planes $P_1: x-y-z-1=0$ and $P_2: 2x+y-3z+4=0$ is given by $P_1 + \lambda P_2 = 0$.$(x-y-z-1) + \lambda(2x+y-3z+4) = 0$$(1+2\lambda)x + (-1+\lambda)y + (-1-3\lambda)z + (-1+4\lambda) = 0$.The distance of this plane from the origin $(0,0,0)$ is given as $\sqrt{\frac{2}{21}}$.The distance formula is $d = \frac{|d_0|}{\sqrt{a^2+b^2+c^2}}$,so:$\frac{|4\lambda-1|}{\sqrt{(1+2\lambda)^2 + (-1+\lambda)^2 + (-1-3\lambda)^2}} = \sqrt{\frac{2}{21}}$.Squaring both sides:$\frac{(4\lambda-1)^2}{(1+4\lambda+4\lambda^2) + (1-2\lambda+\lambda^2) + (1+6\lambda+9\lambda^2)} = \frac{2}{21}$.$\frac{(4\lambda-1)^2}{14\lambda^2+8\lambda+3} = \frac{2}{21}$.$21(16\lambda^2-8\lambda+1) = 2(14\lambda^2+8\lambda+3)$.$336\lambda^2 - 168\lambda + 21 = 28\lambda^2 + 16\lambda + 6$.$308\lambda^2 - 184\lambda + 15 = 0$.Solving the quadratic equation: $308\lambda^2 - 154\lambda - 30\lambda + 15 = 0$.$154\lambda(2\lambda-1) - 15(2\lambda-1) = 0$.$(154\lambda-15)(2\lambda-1) = 0$.Thus,$\lambda = \frac{1}{2}$ or $\lambda = \frac{15}{154}$.For $\lambda = \frac{1}{2}$,the equation is $(x-y-z-1) + \frac{1}{2}(2x+y-3z+4) = 0$.$2x-2y-2z-2 + 2x+y-3z+4 = 0$.$4x-y-5z+2 = 0$.

The equation of a plane at a distance of $\sqrt{\frac{2}{21}}$ from the origin,which contains the line of intersection of the planes $x-y-z-1=0$ and $2x+y-3z+4=0$,is:

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