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(B) The base $OPQR$ is a square in the $xy$-plane with $O(0,0,0)$,$P(3,0,0)$,$Q(3,3,0)$,and $R(0,3,0)$.The mid-point $T$ of $OQ$ is $(\frac{3+0}{2}, \

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$B, C, D$

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(B) The base $OPQR$ is a square in the $xy$-plane with $O(0,0,0)$,$P(3,0,0)$,$Q(3,3,0)$,and $R(0,3,0)$.The mid-point $T$ of $OQ$ is $(\frac{3+0}{2}, \frac{3+0}{2}, 0) = (1.5, 1.5, 0)$.Since $S$ is directly above $T$ with $TS=3$,the coordinates of $S$ are $(1.5, 1.5, 3)$.$(A)$ Vector $\vec{OQ} = 3\hat{i} + 3\hat{j}$ and $\vec{OS} = 1.5\hat{i} + 1.5\hat{j} + 3\hat{k}$.$|\vec{OQ}| = \sqrt{3^2 + 3^2} = 3\sqrt{2}$. $|\vec{OS}| = \sqrt{1.5^2 + 1.5^2 + 3^2} = \sqrt{2.25 + 2.25 + 9} = \sqrt{13.5} = \sqrt{\frac{27}{2}} = \frac{3\sqrt{3}}{\sqrt{2}}$.$\vec{OQ} \cdot \vec{OS} = (3)(1.5) + (3)(1.5) + (0)(3) = 4.5 + 4.5 = 9$.$\cos 	heta = \frac{9}{(3\sqrt{2})(\frac{3\sqrt{3}}{\sqrt{2}})} = \frac{9}{9\sqrt{3}} = \frac{1}{\sqrt{3}}$. Thus,$	heta 
eq \frac{\pi}{3}$. ($A$ is incorrect).$(B)$ The plane containing $	riangle OQS$ passes through $O(0,0,0)$,$Q(3,3,0)$,and $S(1.5, 1.5, 3)$.Normal vector $\vec{n} = \vec{OQ} 	imes \vec{OS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 3 & 0 \ 1.5 & 1.5 & 3 \end{vmatrix} = \hat{i}(9) - \hat{j}(9) + \hat{k}(0) = 9(\hat{i} - \hat{j})$.The equation is $1(x-0) - 1(y-0) + 0(z-0) = 0$,i.e.,$x-y=0$. ($B$ is correct).$(C)$ The perpendicular distance from $P(3,0,0)$ to $x-y=0$ is $d = \frac{|3-0|}{\sqrt{1^2 + (-1)^2}} = \frac{3}{\sqrt{2}}$. ($C$ is correct).$(D)$ Line $RS$ passes through $R(0,3,0)$ and $S(1.5, 1.5, 3)$. Direction vector $\vec{v} = S-R = (1.5, -1.5, 3) = 1.5(\hat{i} - \hat{j} + 2\hat{k})$.Line equation: $\vec{r} = (0,3,0) + \lambda(1, -1, 2)$.Distance from $O(0,0,0)$ to line is $\frac{|\vec{OR} 	imes \vec{v}|}{|\vec{v}|} = \frac{|(0,3,0) 	imes (1,-1,2)|}{|(1,-1,2)|} = \frac{|(6, 0, -3)|}{\sqrt{1+1+4}} = \frac{\sqrt{36+9}}{\sqrt{6}} = \sqrt{\frac{45}{6}} = \sqrt{\frac{15}{2}}$. ($D$ is correct).

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