The equation of the planes parallel to the pl | vedclass

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(A) The equation of any plane parallel to $x - 2y + 2z - 3 = 0$ is of the form $x - 2y + 2z + \lambda = 0$.The distance from the point $(1, 2, 3)$ to

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$4$

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(A) The equation of any plane parallel to $x - 2y + 2z - 3 = 0$ is of the form $x - 2y + 2z + \lambda = 0$.The distance from the point $(1, 2, 3)$ to the plane $x - 2y + 2z + \lambda = 0$ is given by $\frac{|1 - 2(2) + 2(3) + \lambda|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$.Simplifying the expression: $\frac{|1 - 4 + 6 + \lambda|}{\sqrt{9}} = 1 \Rightarrow \frac{|\lambda + 3|}{3} = 1$.This gives $|\lambda + 3| = 3$,so $\lambda + 3 = 3$ or $\lambda + 3 = -3$.Thus,$\lambda = 0$ or $\lambda = -6$.The two possible planes are $x - 2y + 2z = 0$ and $x - 2y + 2z - 6 = 0$.Case $1$: $a=1, b=-2, c=2, d=0$. Then $(b - d) = -2 - 0 = -2$ and $(c - a) = 2 - 1 = 1$. So,$-2 = K(1) \Rightarrow K = -2$.Case $2$: $a=1, b=-2, c=2, d=-6$. Then $(b - d) = -2 - (-6) = 4$ and $(c - a) = 2 - 1 = 1$. So,$4 = K(1) \Rightarrow K = 4$.The positive value of $K$ is $4$.

The equation of the planes parallel to the plane $x - 2y + 2z - 3 = 0$ which are at a unit distance from the point $(1, 2, 3)$ is $ax + by + cz + d = 0$. If $(b - d) = K(c - a)$,then the positive value of $K$ is

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