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Step-by-Step Solution
Step 1: Express the Given Vectors
• Let the position vectors be
$ \overrightarrow{OP} = x\,\hat{i} + y\,\hat{j} - \hat{k} $,
$ \overrightarrow{OQ} = -\hat{i} + 2\,\hat{j} + 3x\,\hat{k} $,
and
$ \overrightarrow{OR} = 3\,\hat{i} + z\,\hat{j} - 7\,\hat{k} $.
• We know that $x, y, z \in \mathbb{R}$, with $x > 0$.
Step 2: Use the Perpendicularity of OP and OQ
For two perpendicular vectors, their dot product is zero. Hence,
$ \overrightarrow{OP} \cdot \overrightarrow{OQ} = 0. $
Compute the dot product:
$ (x\,\hat{i} + y\,\hat{j} - \hat{k}) \cdot (-\hat{i} + 2\,\hat{j} + 3x\,\hat{k}) = -x + 2y - 3x. $
So,
$ -x + 2y - 3x = 0 \quad \Rightarrow \quad -4x + 2y = 0 \quad \Rightarrow \quad 2y = 4x \quad \Rightarrow \quad y = 2x. $
Step 3: Express PQ and Use the Magnitude Condition
We have
$ \overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}
= \bigl(-\hat{i} - x\,\hat{i}\bigr) + \bigl(2\,\hat{j} - y\,\hat{j}\bigr) + \bigl(3x\,\hat{k} + \hat{k}\bigr). $
So,
$ \overrightarrow{PQ} = ( -1 - x )\,\hat{i} + (2 - y)\,\hat{j} + (3x + 1)\,\hat{k}. $
We are given $ |\overrightarrow{PQ}| = \sqrt{20}. $
Thus,
$ \sqrt{20}
= \sqrt{(-1 - x)^2 + (2 - y)^2 + (3x + 1)^2}. $
Squaring both sides gives
$ 20 = (-1 - x)^2 + (2 - y)^2 + (3x + 1)^2. $
Substitute $y = 2x$ into this expression and simplify:
1. $(-1 - x)^2 = 1 + 2x + x^2,$
2. $(2 - y)^2 = (2 - 2x)^2 = 4 - 8x + 4x^2,$
3. $(3x + 1)^2 = 9x^2 + 6x + 1.$
So
$
20
= (1 + 2x + x^2) + (4 - 8x + 4x^2) + (9x^2 + 6x + 1).
$
$
20
= (x^2 + 4x^2 + 9x^2) + (2x - 8x + 6x) + (1 + 4 + 1).
$
$
20
= 14x^2 + 0x + 6.
$
$
14 = 14x^2
\quad \Rightarrow \quad
x^2 = 1
\quad \Rightarrow \quad
x = 1
\;(\text{since } x > 0).
$
Hence, $y = 2x = 2 \cdot 1 = 2.$
Step 4: Apply the Coplanarity Condition
Vectors $ \overrightarrow{OP}, \overrightarrow{OQ}, \text{ and } \overrightarrow{OR} $ are coplanar if their scalar triple product is zero:
$
\bigl(\overrightarrow{OP} \times \overrightarrow{OQ}\bigr) \cdot \overrightarrow{OR} = 0
\quad \text{or equivalently} \quad
\left| \begin{matrix}
x & y & -1 \\
-1 & 2 & 3x \\
3 & z & -7
\end{matrix} \right| = 0.
$
Substitute $ x = 1, y = 2 $ into the determinant:
$
\left| \begin{matrix}
1 & 2 & -1 \\
-1 & 2 & 3 \\
3 & z & -7
\end{matrix} \right| = 0.
$
Compute this determinant. Expand along the first row:
$
1 \bigl[(2)(-7) - (3)(z)\bigr]
- 2 \bigl[(-1)(-7) - (3)(3)\bigr]
- 1 \bigl[(-1)(z) - (2)(3)\bigr] = 0.
$
Simplify each part:
• First part: $1 \times [-14 - 3z] = -14 - 3z.$
• Second part: $-2 \times [7 - 9] = -2 \times (-2) = 4.$
• Third part: $-1 \times [ -z - 6 ] = z + 6.$
So the sum:
$
(-14 - 3z) + 4 + (z + 6) = 0
\quad \Rightarrow \quad
-14 - 3z + 4 + z + 6 = 0.
$
$
-14 + 4 + 6 + (-3z + z) = 0
\quad \Rightarrow \quad
-4 - 2z = 0
\quad \Rightarrow \quad
2z = -4
\quad \Rightarrow \quad
z = -2.
$
Step 5: Find $x^2 + y^2 + z^2$
We have found:
$x = 1, \quad y = 2, \quad z = -2.$
Now,
$
x^2 + y^2 + z^2
= 1^2 + 2^2 + (-2)^2
= 1 + 4 + 4
= 9.
$
Therefore, the required value is
$\boxed{9}.$
