Let the plane pi pass through the point (1,0, | vedclass

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(B) The normal vector $\vec{n}$ of the plane $\pi$ is perpendicular to the normal vectors of the given planes $\vec{n}_1 = (2,3,-1)$ and $\vec{n}_2 =

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(B) The normal vector $\vec{n}$ of the plane $\pi$ is perpendicular to the normal vectors of the given planes $\vec{n}_1 = (2,3,-1)$ and $\vec{n}_2 = (1,-1,2)$.$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(4+1) + \hat{k}(-2-3) = 5\hat{i} - 5\hat{j} - 5\hat{k}$.We can take the normal vector as $\vec{n} = (1, -1, -1)$.The equation of the plane $\pi$ passing through $(1,0,1)$ is $1(x-1) - 1(y-0) - 1(z-1) = 0$,which simplifies to $x-y-z=0$.The plane parallel to $\pi$ passing through $(11,7,5)$ is $x-y-z = k$. Substituting the point $(11,7,5)$,we get $11-7-5 = k$,so $k = -1$.The equation is $x-y-z = -1$,or $x-y-z+1=0$.Comparing this with $ax+by-z-d=0$,we have $a=1, b=-1, d=-1$.Then,$\frac{a}{b} + \frac{b}{d} = \frac{1}{-1} + \frac{-1}{-1} = -1 + 1 = 0$.

Let the plane $\pi$ pass through the point $(1,0,1)$ and be perpendicular to the planes $2x+3y-z=2$ and $x-y+2z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $ax+by-z-d=0$. Then,$\frac{a}{b}+\frac{b}{d}=$

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