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(A) The equation of the plane passing through the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$ is given by:$(2x -

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$\frac{1}{3\sqrt{2}}$

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(A) The equation of the plane passing through the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$ is given by:$(2x - 2y + 3z - 2) + \lambda(x - y + z + 1) = 0$$x(\lambda + 2) - y(\lambda + 2) + z(\lambda + 3) + (\lambda - 2) = 0 \quad \dots(1)$Since this plane contains the line ${L_2}$ formed by the intersection of $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,the normal vector of the plane must be perpendicular to the direction vectors of the lines forming ${L_2}$. Alternatively,the plane must satisfy the condition that the determinant of the coefficients of the three planes is zero:$\begin{vmatrix} \lambda + 2 & -(\lambda + 2) & \lambda + 3 \ 1 & 2 & -1 \ 3 & -1 & 2 \end{vmatrix} = 0$Expanding the determinant:$(\lambda + 2)(4 - 1) + (\lambda + 2)(2 + 3) + (\lambda + 3)(-1 - 6) = 0$$3(\lambda + 2) + 5(\lambda + 2) - 7(\lambda + 3) = 0$$8\lambda + 16 - 7\lambda - 21 = 0 \Rightarrow \lambda = 5$Substituting $\lambda = 5$ into equation $(1)$:$x(5 + 2) - y(5 + 2) + z(5 + 3) + (5 - 2) = 0$$7x - 7y + 8z + 3 = 0$The perpendicular distance from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.$d = \frac{|3|}{\sqrt{7^2 + (-7)^2 + 8^2}} = \frac{3}{\sqrt{49 + 49 + 64}} = \frac{3}{\sqrt{162}} = \frac{3}{9\sqrt{2}} = \frac{1}{3\sqrt{2}}$.

If ${L_1}$ is the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$,and ${L_2}$ is the line of intersection of the planes $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,then the distance of the origin from the plane containing the lines ${L_1}$ and ${L_2}$ is:

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