Let O be the origin. Let overline{OP} = xhat{ | vedclass

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(B) Given $\overline{OP} \perp \overline{OQ}$,their dot product is zero:$(x\hat{i} + y\hat{j} - \hat{k}) \cdot (-\hat{i} + 2\hat{j} + 3x\hat{k}) = 0$$

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$9$

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(B) Given $\overline{OP} \perp \overline{OQ}$,their dot product is zero:$(x\hat{i} + y\hat{j} - \hat{k}) \cdot (-\hat{i} + 2\hat{j} + 3x\hat{k}) = 0$$-x + 2y - 3x = 0 \Rightarrow 2y = 4x \Rightarrow y = 2x \dots (i)$Given $|\overline{PQ}| = \sqrt{20}$,so $|\overline{PQ}|^2 = 20$:$\overline{PQ} = \overline{OQ} - \overline{OP} = (-1-x)\hat{i} + (2-y)\hat{j} + (3x+1)\hat{k}$$|\overline{PQ}|^2 = (-1-x)^2 + (2-y)^2 + (3x+1)^2 = 20$Substitute $y=2x$:$(x+1)^2 + (2-2x)^2 + (3x+1)^2 = 20$$x^2 + 2x + 1 + 4 - 8x + 4x^2 + 9x^2 + 6x + 1 = 20$$14x^2 + 6 = 20 \Rightarrow 14x^2 = 14 \Rightarrow x^2 = 1 \Rightarrow x = 1$ (since $x > 0$)Then $y = 2(1) = 2$.Since $\overline{OP}, \overline{OQ}, \overline{OR}$ are coplanar,their scalar triple product is zero:$\begin{vmatrix} x & y & -1 \ -1 & 2 & 3x \ 3 & z & -7 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} 1 & 2 & -1 \ -1 & 2 & 3 \ 3 & z & -7 \end{vmatrix} = 0$$1(-14 - 3z) - 2(7 - 9) - 1(-z - 6) = 0$$-14 - 3z + 4 + z + 6 = 0 \Rightarrow -2z - 4 = 0 \Rightarrow z = -2$Therefore,$x^2 + y^2 + z^2 = 1^2 + 2^2 + (-2)^2 = 1 + 4 + 4 = 9$.

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