The distance of the point $$(1, -2, 3)$$ from the plane $$x - y + z = 5$$ measured parallel to a line, whose direction ratios are $$2, 3, -6$$, is

We are asked to find the distance of the point $$P(1,\,-2,\,3)$$ from the plane $$x - y + z = 5$$, but the distance must be measured parallel to the line whose direction-ratios are $$2,\;3,\;-6$$. In other words, we must move from the given point in a direction parallel to the vector $$\vec{d}=2\hat i+3\hat j-6\hat k$$ until we hit the plane, and then compute the length of that segment.

First we write the symmetric (parametric) equations of the line that passes through the given point and is parallel to $$\vec d$$. If we take the parameter to be $$t$$, the line through $$P$$ has

$$x = 1 + 2t,\qquad y = -2 + 3t,\qquad z = 3 - 6t.$$

Any point on this line will satisfy these three relations. To locate the particular point at which the line meets the plane, we substitute these coordinates into the plane equation $$x - y + z = 5$$.

So we have

$$\bigl(1 + 2t\bigr) \;-\;\bigl(-2 + 3t\bigr)\;+\;\bigl(3 - 6t\bigr) = 5.$$

Now we carry out the algebra term by term:

$$1 + 2t + 2 - 3t + 3 - 6t = 5.$$ This gathers the constant terms: $$1 + 2 + 3 = 6,$$ and the coefficients of $$t$$: $$2t - 3t - 6t = -7t.$$ So the entire left-hand side becomes $$6 - 7t.$$

Equating to the right-hand side gives

$$6 - 7t = 5.$$

Solving for $$t$$, we subtract 6 from both sides:

$$-7t = 5 - 6 = -1,$$

and then divide by $$-7$$:

$$t = \frac{-1}{-7} = \frac{1}{7}.$$

Thus the required point of intersection, call it $$Q$$, is obtained by substituting $$t = \dfrac17$$ back into the parametric equations:

$$x_Q = 1 + 2\left(\frac17\right) = 1 + \frac27 = \frac97,$$ $$y_Q = -2 + 3\left(\frac17\right) = -2 + \frac37 = -\frac{14}{7} + \frac37 = -\frac{11}{7},$$ $$z_Q = 3 - 6\left(\frac17\right) = 3 - \frac{6}{7} = \frac{21}{7} - \frac{6}{7} = \frac{15}{7}.$$

Now we need the distance between the original point $$P(1,\,-2,\,3)$$ and the intersection point $$Q\!\left(\dfrac97,\,-\dfrac{11}{7},\,\dfrac{15}{7}\right).$$ The displacement vector $$\overrightarrow{PQ}$$ is obtained by subtracting the coordinates of $$P$$ from those of $$Q$$:

$$\overrightarrow{PQ} = \left(\frac97 - 1,\; -\frac{11}{7} - (-2),\; \frac{15}{7} - 3\right).$$

Carrying out each subtraction:

$$\frac97 - 1 = \frac97 - \frac77 = \frac{2}{7},$$ $$-\frac{11}{7} + 2 = -\frac{11}{7} + \frac{14}{7} = \frac{3}{7},$$ $$\frac{15}{7} - 3 = \frac{15}{7} - \frac{21}{7} = -\frac{6}{7}.$$

Hence

$$\overrightarrow{PQ} = \left(\frac{2}{7},\;\frac{3}{7},\;-\frac{6}{7}\right).$$

The distance we seek is simply the magnitude of this vector. Using the three-dimensional distance (or magnitude) formula $$|\vec v| = \sqrt{v_x^{\,2} + v_y^{\,2} + v_z^{\,2}},$$ we get

$$|\overrightarrow{PQ}| = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2} = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1.$$

Therefore, the required distance is $$1$$ unit.

Hence, the correct answer is Option D.