A plane passes through the points $$A(1, 2, 3)$$, $$B(2, 3, 1)$$ and $$C(2, 4, 2)$$. If $$O$$ is the origin and $$P$$ is $$(2, -1, 1)$$, then the projection of $$\vec{OP}$$ on this plane is of length:

We need the equation of the plane passing through $$A(1, 2, 3)$$, $$B(2, 3, 1)$$, and $$C(2, 4, 2)$$. We find two direction vectors in the plane: $$\vec{AB} = (1, 1, -2)$$ and $$\vec{AC} = (1, 2, -1)$$.

The normal to the plane is $$\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}(1 \cdot(-1) - (-2)\cdot 2) - \hat{j}(1\cdot(-1) - (-2)\cdot 1) + \hat{k}(1\cdot 2 - 1\cdot 1) = \hat{i}(-1+4) - \hat{j}(-1+2) + \hat{k}(2-1) = (3, -1, 1)$$.

The equation of the plane is $$3(x-1) - 1(y-2) + 1(z-3) = 0$$, which simplifies to $$3x - y + z = 4$$.

Given $$O$$ is the origin and $$P = (2, -1, 1)$$, we have $$\vec{OP} = (2, -1, 1)$$. The magnitude of $$\vec{OP}$$ is $$|\vec{OP}| = \sqrt{4 + 1 + 1} = \sqrt{6}$$.

The projection of $$\vec{OP}$$ along the normal direction is $$\frac{|\vec{OP} \cdot \vec{n}|}{|\vec{n}|} = \frac{|3(2) + (-1)(-1) + 1(1)|}{\sqrt{9+1+1}} = \frac{|6+1+1|}{\sqrt{11}} = \frac{8}{\sqrt{11}}$$.

The length of the projection of $$\vec{OP}$$ on the plane is given by $$\sqrt{|\vec{OP}|^2 - \left(\frac{\vec{OP} \cdot \vec{n}}{|\vec{n}|}\right)^2} = \sqrt{6 - \frac{64}{11}} = \sqrt{\frac{66 - 64}{11}} = \sqrt{\frac{2}{11}}$$.

Therefore, the projection of $$\vec{OP}$$ on the plane is of length $$\sqrt{\dfrac{2}{11}}$$.