The given plane is 2x + 3y – z = 5 ...(1)Direction ratios of the normal to the plane are < 2, 3, – 1 >.∴ Direction cosines of the normal to the plane are :$\langle \frac{2}{\sqrt{4+9+1}}, \frac{3}{\sqrt{4+9+1}}, \frac{-1}{\sqrt{4+9+1}} \rangle$i.e., $\langle \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \rangle$And distance of (1) from the origin $= \frac{\left|0+0-0-5\right|}{\sqrt{14}} = \frac{5}{\sqrt{14}}$