Key Concepts and Formulas

• Dot Product: Two vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if their dot product is zero: $\vec{a} \cdot \vec{b} = 0$. If $\vec{a} = a_1\widehat i + a_2\widehat j + a_3\widehat k$ and $\vec{b} = b_1\widehat i + b_2\widehat j + b_3\widehat k$, then $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$.
• Vector Subtraction and Magnitude: The vector from point P to point Q is $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$. The magnitude of a vector $\vec{v} = v_x\widehat i + v_y\widehat j + v_z\widehat k$ is $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.
• Scalar Triple Product (Coplanarity): Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero: $[\vec{a}, \vec{b}, \vec{c}] = 0$. This can be computed as the determinant of the matrix formed by their components.

Step-by-Step Solution

Step 1: Utilize the perpendicularity condition. We are given that $\overrightarrow{OP}$ is perpendicular to $\overrightarrow{OQ}$. This means their dot product is zero. $\overrightarrow{OP} \cdot \overrightarrow{OQ} = (x\widehat i + y\widehat j - \widehat k) \cdot (- \widehat i + 2\widehat j + 3x\widehat k) = 0$ $x(-1) + y(2) + (-1)(3x) = 0$ $-x + 2y - 3x = 0$ $-4x + 2y = 0$ Dividing by 2, we get a relationship between $x$ and $y$: $2y = 4x \implies y = 2x \quad (*)$

Step 2: Utilize the magnitude condition. We are given that $|\overrightarrow{PQ}| = \sqrt{20}$. First, we find the vector $\overrightarrow{PQ}$. $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$ $\overrightarrow{PQ} = (- \widehat i + 2\widehat j + 3x\widehat k) - (x\widehat i + y\widehat j - \widehat k)$ $\overrightarrow{PQ} = (-1 - x)\widehat i + (2 - y)\widehat j + (3x + 1)\widehat k$ Now, we use the magnitude condition: $|\overrightarrow{PQ}|^2 = 20$. $(-1 - x)^2 + (2 - y)^2 + (3x + 1)^2 = 20$ $(1 + x)^2 + (2 - y)^2 + (3x + 1)^2 = 20$ Substitute $y = 2x$ from equation $(*)$: $(1 + x)^2 + (2 - 2x)^2 + (3x + 1)^2 = 20$ Expand the terms: $(1 + 2x + x^2) + (4 - 8x + 4x^2) + (9x^2 + 6x + 1) = 20$ Combine like terms: $(x^2 + 4x^2 + 9x^2) + (2x - 8x + 6x) + (1 + 4 + 1) = 20$ $14x^2 + 0x + 6 = 20$ $14x^2 = 14$ $x^2 = 1$ Since we are given $x > 0$, we have $x = 1$. Now, find $y$ using $y = 2x$: $y = 2(1) \implies y = 2$

Step 3: Utilize the coplanarity condition. We are given that $\overrightarrow{OR}$ is coplanar with $\overrightarrow{OP}$ and $\overrightarrow{OQ}$. This means their scalar triple product is zero. The vectors are: $\overrightarrow{OP} = x\widehat i + y\widehat j - \widehat k = 1\widehat i + 2\widehat j - \widehat k$ $\overrightarrow{OQ} = - \widehat i + 2\widehat j + 3x\widehat k = - \widehat i + 2\widehat j + 3(1)\widehat k = - \widehat i + 2\widehat j + 3\widehat k$ $\overrightarrow{OR} = 3\widehat i + z\widehat j - 7\widehat k$ The scalar triple product is given by the determinant: $[\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{OR}] = \begin{vmatrix} 1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7 \end{vmatrix} = 0$ Expand the determinant: $1 \begin{vmatrix} 2 & 3 \\ z & -7 \end{vmatrix} - 2 \begin{vmatrix} -1 & 3 \\ 3 & -7 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 2 \\ 3 & z \end{vmatrix} = 0$ $1((2)(-7) - (3)(z)) - 2((-1)(-7) - (3)(3)) - 1((-1)(z) - (2)(3)) = 0$ $1(-14 - 3z) - 2(7 - 9) - 1(-z - 6) = 0$ $-14 - 3z - 2(-2) + z + 6 = 0$ $-14 - 3z + 4 + z + 6 = 0$ Combine terms: $(-14 + 4 + 6) + (-3z + z) = 0$ $-4 - 2z = 0$ $-2z = 4 \implies z = -2$

Step 4: Calculate the final expression. We need to find the value of $x^2 + y^2 + z^2$. We have $x = 1$, $y = 2$, and $z = -2$. $x^2 + y^2 + z^2 = (1)^2 + (2)^2 + (-2)^2$ $x^2 + y^2 + z^2 = 1 + 4 + 4$ $x^2 + y^2 + z^2 = 9$

Common Mistakes & Tips

• Sign Errors in Determinants: Be extremely careful with signs when expanding determinants, especially when dealing with negative components. Double-checking each term can prevent errors.
• Algebraic Simplification: Always simplify equations as much as possible. For instance, $y=2x$ is much easier to work with than $-4x+2y=0$.
• Using the Constraint $x>0$: Remember to use all given constraints. The condition $x>0$ is crucial for selecting the correct value of $x$ from $x^2=1$.

Summary

The problem was solved by systematically applying the conditions of vector perpendicularity, vector magnitude, and vector coplanarity. The perpendicularity condition gave a relationship between $x$ and $y$. The magnitude condition, along with the derived relationship, allowed us to solve for $x$ and then $y$. Finally, the coplanarity condition was used to find the value of $z$. Substituting these values into the expression $x^2 + y^2 + z^2$ yielded the final answer.

The final answer is $\boxed{9}$.