The plane containing the line $$\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z-1}{3}$$ and also containing its projection on the plane $$2x + 3y - z = 5$$, contains which one of the following points?

We have the given straight line in symmetric form

$$\frac{x-3}{2}\;=\;\frac{y+2}{-1}\;=\;\frac{z-1}{3}$$

To change it into the parametric form, let the common ratio be $$t$$. Thus

$$x = 3 + 2t,\qquad y = -2 - t,\qquad z = 1 + 3t.$$

Accordingly, a fixed point on the line is

$$P_0(3,-2,1)$$

and its direction vector is

$$\vec d = \langle 2,\,-1,\,3\rangle.$$

The other surface mentioned is the plane

$$2x + 3y - z = 5,$$

whose normal vector is

$$\vec n = \langle 2,\,3,\,-1\rangle.$$

The required plane must contain the original line and also the projection of that line on the plane $$2x+3y-z=5.$$ To construct the projection, we need a direction vector that lies entirely inside the plane. The well-known formula for projecting a vector $$\vec d$$ onto a plane with normal $$\vec n$$ is

$$\vec d_{\text{proj}} = \vec d \;-\;\frac{\vec d\cdot\vec n}{|\vec n|^{2}}\;\vec n.$$

First compute the dot product:

$$\vec d\cdot\vec n = 2(2) + (-1)(3) + 3(-1) = 4 - 3 - 3 = -2.$$

The squared length of the normal is

$$|\vec n|^{2} = 2^{2} + 3^{2} + (-1)^{2} = 4 + 9 + 1 = 14.$$

Hence

$$\frac{\vec d\cdot\vec n}{|\vec n|^{2}} = \frac{-2}{14} = -\frac{1}{7}.$$

Substituting into the projection formula,

$$\vec d_{\text{proj}} = \vec d -\!\left(-\frac{1}{7}\right)\vec n = \vec d + \frac{1}{7}\,\vec n.$$

That is,

$$\vec d_{\text{proj}} = \Big\langle 2,\,-1,\,3\Big\rangle + \frac{1}{7}\Big\langle 2,\,3,\,-1\Big\rangle = \Big\langle 2+\tfrac{2}{7},\,-1+\tfrac{3}{7},\,3-\tfrac{1}{7}\Big\rangle = \Big\langle \tfrac{16}{7},\,-\tfrac{4}{7},\,\tfrac{20}{7}\Big\rangle.$$

Multiplying by the common denominator 7 we may take the simpler, parallel vector

$$\vec v = \langle 16,\,-4,\,20\rangle = 4\langle 4,\,-1,\,5\rangle.$$

Thus we adopt

$$\vec v = \langle 4,\,-1,\,5\rangle$$

as the direction vector of the projected line.

Next we need a specific point on the projection itself. The projection of the point $$P_0(3,-2,1)$$ onto the plane is obtained by dropping a perpendicular along $$\vec n$$. A standard result is

$$\text{Proj}_{\text{plane}}(A) = A - \lambda\,\vec n,$$

where

$$\lambda = \frac{\vec n\cdot A - 5}{|\vec n|^{2}},$$

because the plane equation is $$\vec n\cdot\vec r = 5.$$

Compute the numerator:

$$\vec n\cdot P_0 - 5 = [2(3) + 3(-2) + (-1)(1)] - 5 = (6 - 6 - 1) - 5 = -1 - 5 = -6.$$

Therefore

$$\lambda = \frac{-6}{14} = -\frac{3}{7}.$$

So the projected point is

$$P' = P_0 - \left(-\frac{3}{7}\right)\vec n = (3,-2,1) + \frac{3}{7}\langle 2,\,3,\,-1\rangle = \Big(3+\tfrac{6}{7},\,-2+\tfrac{9}{7},\,1-\tfrac{3}{7}\Big) = \Big(\tfrac{27}{7},\,-\tfrac{5}{7},\,\tfrac{4}{7}\Big).$$

We now possess two non-parallel direction vectors contained in the desired plane, namely

$$\vec d = \langle 2,\,-1,\,3\rangle$$

and

$$\vec v = \langle 4,\,-1,\,5\rangle,$$

and a point $$P_0(3,-2,1)$$ on that plane. The normal to the required plane is provided by their cross product:

$$\vec n_{\text{plane}} = \vec d \times \vec v.$$

Carrying out the determinant expansion,

$$\vec n_{\text{plane}} = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 2 & -1 & 3\\ 4 & -1 & 5 \end{vmatrix} = \mathbf i\big((-1)(5) - 3(-1)\big)\;-\; \mathbf j\big(2\cdot5 - 3\cdot4\big)\;+\; \mathbf k\big(2(-1) - (-1)4\big)$$ $$= \mathbf i(-5 + 3) - \mathbf j(10 - 12) + \mathbf k(-2 + 4)$$ $$= -2\,\mathbf i + 2\,\mathbf j + 2\,\mathbf k.$$

A convenient proportional form is

$$\vec n_{\text{plane}} = \langle -2,\,2,\,2\rangle = -2\langle 1,\,-1,\,-1\rangle,$$

so we may take the simpler normal

$$\langle 1,\,-1,\,-1\rangle.$$

The scalar equation of a plane through $$P_0(3,-2,1)$$ with this normal is

$$(x-3)\cdot1 + (y+2)(-1) + (z-1)(-1) = 0.$$

Simplifying term by term,

$$x - 3 - y - 2 - z + 1 = 0,$$

which yields

$$x - y - z - 4 = 0,$$

or more neatly,

$$x - y - z = 4.$$

All points contained in the required plane must satisfy this linear equation. We now test each option:

A. For (2, 2, 0): $$2 - 2 - 0 = 0\neq 4.$$(False)

B. For (−2, 2, 2): $$-2 - 2 - 2 = -6\neq 4.$$(False)

C. For (0, −2, 2): $$0 - (-2) - 2 = 0\neq 4.$$(False)

D. For (2, 0, −2): $$2 - 0 - (-2) = 2 + 2 = 4\,,$$ which indeed satisfies the equation. (True)

Hence, the correct answer is Option D.