Let the position vectors of points 'A' and 'B' be $$\hat{i} + \hat{j} + \hat{k}$$ and $$2\hat{i} + \hat{j} + 3\hat{k}$$, respectively. A point 'P' divides the line segment AB internally in the ratio $$\lambda : 1$$ $$(\lambda > 0)$$. If O is the origin and $$\vec{OB} \cdot \vec{OP} - 3|\vec{OA} \times \vec{OP}|^2 = 6$$ then $$\lambda$$ is equal to ___________.

We have the position vectors of the two fixed points

$$\vec{OA}=1\hat i+1\hat j+1\hat k=(1,1,1),\qquad \vec{OB}=2\hat i+1\hat j+3\hat k=(2,1,3).$$

A point $$P$$ divides the line segment $$AB$$ internally in the ratio $$\lambda:1\;(\lambda>0).$$ For internal division, the section-formula tells us

$$\vec{OP}=\frac{\lambda\,\vec{B}+1\,\vec{A}}{\lambda+1}.$$

Substituting $$\vec{A}=(1,1,1)$$ and $$\vec{B}=(2,1,3)$$ we obtain

$$\vec{OP}= \frac{\lambda(2,1,3)+(1,1,1)}{\lambda+1} =\frac{(2\lambda+1,\;\lambda+1,\;3\lambda+1)}{\lambda+1} =\left(\frac{2\lambda+1}{\lambda+1},\;1,\;\frac{3\lambda+1}{\lambda+1}\right).$$

Now we evaluate the two vector expressions that appear in the given condition.

1. The dot product $$\vec{OB}\cdot\vec{OP}$$

$$\vec{OB}\cdot\vec{OP} =2\left(\frac{2\lambda+1}{\lambda+1}\right) +1\left(1\right) +3\left(\frac{3\lambda+1}{\lambda+1}\right) =\frac{4\lambda+2+\lambda+1+9\lambda+3}{\lambda+1} =\frac{14\lambda+6}{\lambda+1}.$$

2. The cross product $$\vec{OA}\times\vec{OP}$$

Writing $$\vec{OA}=(1,1,1)$$ and $$\vec{OP}=\left(\dfrac{2\lambda+1}{\lambda+1},\,1,\,\dfrac{3\lambda+1}{\lambda+1}\right)$$, the determinant form gives

$$\vec{OA}\times\vec{OP} =\Bigl(\,OP_z-1,\;OP_x-OP_z,\;1-OP_x\Bigr).$$

Because $$OP_x=\dfrac{2\lambda+1}{\lambda+1}$$ and $$OP_z=\dfrac{3\lambda+1}{\lambda+1},$$ we have

$$OP_z-1=\frac{3\lambda+1}{\lambda+1}-1 =\frac{2\lambda}{\lambda+1},$$ $$OP_x-OP_z=\frac{2\lambda+1}{\lambda+1}-\frac{3\lambda+1}{\lambda+1} =-\frac{\lambda}{\lambda+1},$$ $$1-OP_x=1-\frac{2\lambda+1}{\lambda+1} =-\frac{\lambda}{\lambda+1}.$$

Thus

$$\vec{OA}\times\vec{OP} =\left(\frac{2\lambda}{\lambda+1},\;-\frac{\lambda}{\lambda+1},\;-\frac{\lambda}{\lambda+1}\right).$$

The squared magnitude required is

$$\bigl|\vec{OA}\times\vec{OP}\bigr|^{2} =\left(\frac{2\lambda}{\lambda+1}\right)^{2} +\left(-\frac{\lambda}{\lambda+1}\right)^{2} +\left(-\frac{\lambda}{\lambda+1}\right)^{2} =\frac{4\lambda^{2}+\lambda^{2}+\lambda^{2}}{(\lambda+1)^{2}} =\frac{6\lambda^{2}}{(\lambda+1)^{2}}.$$

According to the problem, the vectors satisfy

$$\vec{OB}\cdot\vec{OP}-3\bigl|\vec{OA}\times\vec{OP}\bigr|^{2}=6.$$

Substituting the two expressions we just found:

$$\frac{14\lambda+6}{\lambda+1}-3\left(\frac{6\lambda^{2}}{(\lambda+1)^{2}}\right)=6.$$

To clear the denominators, multiply every term by $$(\lambda+1)^{2}$$:

$$\bigl(14\lambda+6\bigr)(\lambda+1)-18\lambda^{2}=6(\lambda+1)^{2}.$$

Expanding each side gives

$$14\lambda^{2}+20\lambda+6-18\lambda^{2}=6\lambda^{2}+12\lambda+6.$$

Combining like terms on the left:

$$-4\lambda^{2}+20\lambda+6 = 6\lambda^{2}+12\lambda+6.$$

Bringing everything to one side:

$$-4\lambda^{2}+20\lambda+6-6\lambda^{2}-12\lambda-6 = 0,$$ $$-10\lambda^{2}+8\lambda = 0.$$

Factoring out $$-2\lambda$$ we get

$$-2\lambda(5\lambda-4)=0.$$

This yields the possibilities $$\lambda=0$$ or $$5\lambda-4=0.$$ Because the ratio must be positive, we reject $$\lambda=0$$ and keep

$$\lambda=\frac{4}{5}=0.8.$$

So, the answer is $$0.8$$.