Let $$O$$ be the origin. Let $$\vec{OP} = x\hat{i} + y\hat{j} - \hat{k}$$ and $$\vec{OQ} = -\hat{i} + 2\hat{j} + 3x\hat{k}$$, $$x, y \in R$$, $$x > 0$$, be such that $$|\vec{PQ}| = \sqrt{20}$$ and the vector $$\vec{OP}$$ is perpendicular to $$\vec{OQ}$$. If $$\vec{OR} = 3\hat{i} + z\hat{j} - 7\hat{k}$$, $$z \in R$$, is coplanar with $$\vec{OP}$$ and $$\vec{OQ}$$, then the value of $$x^2 + y^2 + z^2$$ is equal to:

We have $$\vec{OP} = x\hat{i} + y\hat{j} - \hat{k}$$ and $$\vec{OQ} = -\hat{i} + 2\hat{j} + 3x\hat{k}$$, with $$x > 0$$.

Condition 1: $$\vec{OP} \perp \vec{OQ}$$, so $$\vec{OP} \cdot \vec{OQ} = 0$$. This gives $$x(-1) + y(2) + (-1)(3x) = 0$$, so $$-x + 2y - 3x = 0$$, which simplifies to $$2y = 4x$$, giving $$y = 2x$$ ... (1).

Condition 2: $$|\vec{PQ}| = \sqrt{20}$$. We have $$\vec{PQ} = \vec{OQ} - \vec{OP} = (-1-x)\hat{i} + (2-y)\hat{j} + (3x+1)\hat{k}$$.

$$|\vec{PQ}|^2 = (-1-x)^2 + (2-y)^2 + (3x+1)^2 = 20$$.

Substituting $$y = 2x$$: $$(1+x)^2 + (2-2x)^2 + (3x+1)^2 = 20$$.

Expanding: $$(1 + 2x + x^2) + (4 - 8x + 4x^2) + (9x^2 + 6x + 1) = 20$$.

$$14x^2 + 0x + 6 = 20$$, so $$14x^2 = 14$$, giving $$x^2 = 1$$, so $$x = 1$$ (since $$x > 0$$).

From (1): $$y = 2$$. So $$\vec{OP} = \hat{i} + 2\hat{j} - \hat{k}$$ and $$\vec{OQ} = -\hat{i} + 2\hat{j} + 3\hat{k}$$.

Now $$\vec{OR} = 3\hat{i} + z\hat{j} - 7\hat{k}$$ is coplanar with $$\vec{OP}$$ and $$\vec{OQ}$$. Three vectors are coplanar when their scalar triple product is zero:

$$\begin{vmatrix} 1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7 \end{vmatrix} = 0$$.

Expanding along the first row: $$1(2 \cdot (-7) - 3z) - 2((-1)(-7) - 3 \cdot 3) + (-1)((-1)z - 2 \cdot 3) = 0$$.

$$1(-14 - 3z) - 2(7 - 9) + (-1)(-z - 6) = 0$$.

$$-14 - 3z - 2(-2) + z + 6 = 0$$.

$$-14 - 3z + 4 + z + 6 = 0$$, so $$-2z - 4 = 0$$, giving $$z = -2$$.

Therefore $$x^2 + y^2 + z^2 = 1 + 4 + 4 = 9$$.

The answer is $$9$$, which is Option B.