Equation of planes parallel to the plane x-2y | vedclass

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(B) The equation of any plane parallel to the plane $x-2y+2z+4=0$ is given by $x-2y+2z+\lambda=0$. The distance $d$ from a point $(x_1, y_1, z_1)$ to

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$x-2y+2z=0, x-2y+2z-6=0$

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(B) The equation of any plane parallel to the plane $x-2y+2z+4=0$ is given by $x-2y+2z+\lambda=0$. The distance $d$ from a point $(x_1, y_1, z_1)$ to the plane $ax+by+cz+d=0$ is given by $d = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$. Given the distance is $1$ unit from the point $(1, 2, 3)$,we have: $\frac{|1(1)-2(2)+2(3)+\lambda|}{\sqrt{1^2+(-2)^2+2^2}} = 1$ $\frac{|1-4+6+\lambda|}{\sqrt{1+4+4}} = 1$ $\frac{|3+\lambda|}{3} = 1$ $|3+\lambda| = 3$ This implies $3+\lambda = 3$ or $3+\lambda = -3$. For $3+\lambda = 3$,we get $\lambda = 0$. For $3+\lambda = -3$,we get $\lambda = -6$. Substituting these values back into the equation $x-2y+2z+\lambda=0$,we get the required planes as $x-2y+2z=0$ and $x-2y+2z-6=0$.

Equation of planes parallel to the plane $x-2y+2z+4=0$ which are at a distance of one unit from the point $(1,2,3)$ are

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