Let the position vectors of the points P, Q, R and S be $$\vec{a} = \hat{i} + 2\hat{j} - 5\hat{k}$$, $$\vec{b} = 3\hat{i} + 6\hat{j} + 3\hat{k}$$, $$\vec{c} = \frac{17}{5}\hat{i} + \frac{16}{5}\hat{j} + 7\hat{k}$$ and $$\vec{d} = 2\hat{i} + \hat{j} + \hat{k}$$, respectively. Then which of the following statements is true?

The position vectors are
$$\vec{a}=1\,\hat{i}+2\,\hat{j}-5\,\hat{k},\; \vec{b}=3\,\hat{i}+6\,\hat{j}+3\,\hat{k},\; \vec{c}=\tfrac{17}{5}\,\hat{i}+\tfrac{16}{5}\,\hat{j}+7\,\hat{k},\; \vec{d}=2\,\hat{i}+1\,\hat{j}+1\,\hat{k}.$$

Step 1: Coplanarity of P, Q, R, S
Four points are coplanar iff the scalar triple product of any three displacement vectors formed with the fourth point is zero.
Choose P as the reference point.
$$\vec{PQ}= \vec{b}-\vec{a}= \left(2,4,8\right),$$ $$\vec{PR}= \vec{c}-\vec{a}= \left(\tfrac{12}{5},\tfrac{6}{5},12\right),$$ $$\vec{PS}= \vec{d}-\vec{a}= \left(1,-1,6\right).$$

To avoid fractions multiply $$\vec{PR}$$ by 5 (scalar multiplication does not affect coplanarity): $$5\vec{PR}=(12,6,60).$$
Compute $$\left[\vec{PQ},\,5\vec{PR},\,\vec{PS}\right] =\begin{vmatrix} 2 & 4 & 8\\ 12 & 6 & 60\\ 1 & -1 & 6 \end{vmatrix}.$$ Evaluating the determinant:
$$2\,(6\cdot6-60\cdot(-1))- 4\,(12\cdot6-60\cdot1)+ 8\,(12\cdot(-1)-6\cdot1)$$ $$=2\,(36+60)-4\,(72-60)+8\,(-12-6)$$ $$=2\cdot96-4\cdot12+8\cdot(-18)=192-48-144=0.$$ Since the scalar triple product is $$0,$$ the points are coplanar. Hence Option A is false.

Step 2: Position vector $$\dfrac{\vec{b}+2\vec{d}}{3}$$
Compute $$\vec{b}+2\vec{d}=(3,6,3)+2(2,1,1)=(3+4,\,6+2,\,3+2)=(7,8,5).$$
Therefore $$\frac{\vec{b}+2\vec{d}}{3}=\left(\frac{7}{3},\frac{8}{3},\frac{5}{3}\right).$$

Step 3: Does this point divide PR in the ratio 5 : 4?
Let the required point divide the segment joining P($$\vec{a}$$) and R($$\vec{c}$$) internally in the ratio 5 : 4 (5 parts towards R, 4 parts towards P).
Using the internal division formula:
$$\vec{OP}=\vec{a},\; \vec{OR}=\vec{c},\; m:n=5:4.$$ Then the position vector of the point is $$\vec{v}=\frac{5\,\vec{c}+4\,\vec{a}}{5+4} =\frac{5\,\vec{c}+4\,\vec{a}}{9}.$$ Compute the numerator:

$$5\vec{c}=5\!\left(\tfrac{17}{5},\tfrac{16}{5},7\right) =(17,16,35),$$ $$4\vec{a}=4(1,2,-5)=(4,8,-20).$$ Hence $$5\vec{c}+4\vec{a}=(17+4,\;16+8,\;35-20)=(21,24,15).$$ Divide by 9: $$\vec{v}=\left(\frac{21}{9},\frac{24}{9},\frac{15}{9}\right) =\left(\frac{7}{3},\frac{8}{3},\frac{5}{3}\right).$$

This is exactly $$\dfrac{\vec{b}+2\vec{d}}{3}$$.
Therefore that vector represents the internal division point of PR in the ratio 5 : 4, confirming Option B as true.

Step 4: External division (Option C)
If the point had divided PR externally in the same ratio, its position vector would be $$\frac{5\,\vec{c}-4\,\vec{a}}{5-4}=5\vec{c}-4\vec{a} =(17,16,35)-(4,8,-20)=(13,8,55),$$ which is not $$\left(\frac{7}{3},\frac{8}{3},\frac{5}{3}\right)$$.
Thus Option C is false.

Step 5: Magnitude of $$\vec{b}\times\vec{d}$$ (Option D)
Compute the cross product:
$$\vec{b}\times\vec{d} =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 6 & 3\\ 2 & 1 & 1 \end{vmatrix} =\hat{i}(6\cdot1-3\cdot1) -\hat{j}(3\cdot1-3\cdot2) +\hat{k}(3\cdot1-6\cdot2)$$ $$=3\,\hat{i}+3\,\hat{j}-9\,\hat{k}.$$ Magnitude squared:
$$|\vec{b}\times\vec{d}|^{2}=3^{2}+3^{2}+(-9)^{2}=9+9+81=99\neq95.$$ Hence Option D is false.

Only Option B is correct.

Final Answer: Option B which is: $$\dfrac{\vec{b}+2\vec{d}}{3}$$ is the position vector of a point which divides PR internally in the ratio 5 : 4.