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(A) The equation of the plane $\pi_1$ passing through $(0,1,2), (1,0,-2), (-2,1,0)$ is given by the determinant:$\begin{vmatrix} x-0 & y-1 & z-2 \ 1-

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$\frac{\sqrt{14}}{9}$

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(A) The equation of the plane $\pi_1$ passing through $(0,1,2), (1,0,-2), (-2,1,0)$ is given by the determinant:$\begin{vmatrix} x-0 & y-1 & z-2 \ 1-0 & 0-1 & -2-2 \ -2-0 & 1-1 & 0-2 \end{vmatrix} = 0$$\Rightarrow \begin{vmatrix} x & y-1 & z-2 \ 1 & -1 & -4 \ -2 & 0 & -2 \end{vmatrix} = 0$Expanding along the first row:$x(2-0) - (y-1)(-2-8) + (z-2)(0-2) = 0$$2x + 10(y-1) - 2(z-2) = 0$$2x + 10y - 10 - 2z + 4 = 0$$2x + 10y - 2z - 6 = 0 \Rightarrow x + 5y - z = 3$. The normal vector is $\vec{n_1} = (1, 5, -1)$.The plane $\pi_2$ passes through $(1,2,3)$ and is perpendicular to $x+y+z=1$ and $2x-3y+z=5$. The normal vector $\vec{n_2}$ is the cross product of the normals $(1,1,1)$ and $(2,-3,1)$:$\vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -3 & 1 \end{vmatrix} = \hat{i}(1+3) - \hat{j}(1-2) + \hat{k}(-3-2) = 4\hat{i} + \hat{j} - 5\hat{k}$.The equation of $\pi_2$ is $4(x-1) + 1(y-2) - 5(z-3) = 0 \Rightarrow 4x + y - 5z + 9 = 0$.The acute angle $	heta$ between $\pi_1$ and $\pi_2$ is given by:$\cos 	heta = \left| \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} ight| = \left| \frac{(1)(4) + (5)(1) + (-1)(-5)}{\sqrt{1^2+5^2+(-1)^2} \sqrt{4^2+1^2+(-5)^2}} ight|$$= \left| \frac{4+5+5}{\sqrt{27} \sqrt{42}} ight| = \frac{14}{\sqrt{9 	imes 3} \sqrt{6 	imes 7}} = \frac{14}{3\sqrt{3} \sqrt{6} \sqrt{7}} = \frac{14}{3 \sqrt{18 	imes 7}} = \frac{14}{3 \sqrt{126}} = \frac{14}{3 	imes 3 \sqrt{14}} = \frac{\sqrt{14}}{9}$.

Let $\pi_1$ be the plane passing through the points $(0,1,2), (1,0,-2), (-2,1,0)$ and $\pi_2$ be the plane passing through the point $(1,2,3)$ and perpendicular to the planes $x+y+z=1$ and $2x-3y+z=5$. If $\theta$ is the acute angle between the planes $\pi_1$ and $\pi_2$, then $\cos \theta=$

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