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(B) The normal vector $\vec{n}$ to the plane is given by the cross product of the two parallel vectors $\vec{v_1} = 2\hat{i}+3\hat{j}+\hat{k}$ and $\v

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$3$

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(B) The normal vector $\vec{n}$ to the plane is given by the cross product of the two parallel vectors $\vec{v_1} = 2\hat{i}+3\hat{j}+\hat{k}$ and $\vec{v_2} = -\hat{i}+2\hat{j}-3\hat{k}$.$\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 1 \ -1 & 2 & -3 \end{vmatrix} = \hat{i}(-9-2) - \hat{j}(-6+1) + \hat{k}(4+3) = -11\hat{i} + 5\hat{j} + 7\hat{k}$.The equation of the plane passing through $(1, 2, 1)$ with normal vector $\vec{n} = -11\hat{i} + 5\hat{j} + 7\hat{k}$ is:$-11(x-1) + 5(y-2) + 7(z-1) = 0$$-11x + 11 + 5y - 10 + 7z - 7 = 0$$-11x + 5y + 7z - 6 = 0$$-11x + 5y + 7z = 6$Dividing by $6$,we get:$-\frac{11}{6}x + \frac{5}{6}y + \frac{7}{6}z = 1$Comparing this with $ax+by+cz=1$,we have $a = -\frac{11}{6}$,$b = \frac{5}{6}$,$c = \frac{7}{6}$.Therefore,$18(a+b+c) = 18 \left(-\frac{11}{6} + \frac{5}{6} + \frac{7}{6}ight) = 18 \left(\frac{1}{6}ight) = 3$.

If the cartesian equation of the plane passing through the point $\hat{i}+2 \hat{j}+\hat{k}$ and parallel to the vectors $2 \hat{i}+3 \hat{j}+\hat{k}$ and $-\hat{i}+2 \hat{j}-3 \hat{k}$ is $a x+b y+c z=1$,then $18(a+b+c)$ is equal to

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