A plane pi given by ax + by + 11z + d = 0 is | vedclass

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(D) The normal vector to the plane $\pi: ax + by + 11z + d = 0$ is $\vec{n} = a\hat{i} + b\hat{j} + 11\hat{k}$.Since $\pi$ is perpendicular to $2x - 3

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$-3ab$

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(D) The normal vector to the plane $\pi: ax + by + 11z + d = 0$ is $\vec{n} = a\hat{i} + b\hat{j} + 11\hat{k}$.Since $\pi$ is perpendicular to $2x - 3y + z = 4$ and $3x + y - z = 5$,$\vec{n}$ is perpendicular to $\vec{n}_1 = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{n}_2 = 3\hat{i} + \hat{j} - \hat{k}$.Thus,$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 1 \ 3 & 1 & -1 \end{vmatrix} = \hat{i}(3-1) - \hat{j}(-2-3) + \hat{k}(2+9) = 2\hat{i} + 5\hat{j} + 11\hat{k}$.Comparing this with $\vec{n} = a\hat{i} + b\hat{j} + 11\hat{k}$,we get $a = 2$ and $b = 5$.The plane equation is $2x + 5y + 11z + d = 0$.The perpendicular distance from $(0,0,0)$ to the plane is $\frac{|d|}{\sqrt{2^2 + 5^2 + 11^2}} = \frac{|d|}{\sqrt{4 + 25 + 121}} = \frac{|d|}{\sqrt{150}} = \frac{|d|}{5\sqrt{6}}$.Given distance is $\sqrt{6}$,so $\frac{|d|}{5\sqrt{6}} = \sqrt{6} \implies |d| = 5 	imes 6 = 30$.Since intercepts are positive,the plane is $2x + 5y + 11z = -d$. The intercepts are $x = -d/2, y = -d/5, z = -d/11$. For these to be positive,$d$ must be negative,so $d = -30$.We have $ab = 2 	imes 5 = 10$. Since $d = -30$,$d = -3ab$.

$A$ plane $\pi$ given by $ax + by + 11z + d = 0$ is perpendicular to the planes $2x - 3y + z = 4$ and $3x + y - z = 5$. The perpendicular distance from the origin to the plane $\pi$ is $\sqrt{6}$ units. If all the intercepts made by the plane $\pi$ on the coordinate axes are positive,then $d =$

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