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

Let A be any point on the plane x - y + z = 5 and B(1, -2, 3).

Then equation of the line AB whose direction ratios are 2, 3, − 6

$\begin{array}{l} \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}-3}{-6}=\mathrm{\lambda }\left(\text{ Let }\right)\\ \Rightarrow \mathrm{x}=1+2\mathrm{\lambda },\mathrm{y}=-2+3\mathrm{\lambda },\mathrm{z}=3-6\mathrm{\lambda }\\ \mathrm{A}(1+2\mathrm{\lambda }, -2+3\mathrm{\lambda }, 3-6\mathrm{\lambda })\end{array}$

A lies on plane.

Then,   $1+2\mathrm{\lambda }-(-2+3\mathrm{\lambda })+3-6\mathrm{\lambda }=5$

$\begin{array}{l}\Rightarrow 1+2\mathrm{\lambda }+2-3\mathrm{\lambda }+3-6\mathrm{\lambda }=5\Rightarrow \mathrm{\lambda }=\frac{1}{7}\\ \therefore \mathrm{A}\left(\frac{9}{7},\frac{-11}{7},\frac{15}{7}\right)\\ \text{ Distance }\mathrm{AB}=\sqrt{{\left(1-\frac{9}{7}\right)}^2+{\left(-2+\frac{11}{7}\right)}^2+{\left(3-\frac{15}{7}\right)}^2}\\ =\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1\end{array}$