Find the angle between the planes $$x+y+z=1$$​ and $$x-2y+3z=1$$.

Correct option is B

​$$\begin{aligned}&\bullet \ \text{Normal vector to Plane 1 } (\mathbf{n}_1): \langle 1, 1, 1 \rangle \\&\bullet \ \text{Normal vector to Plane 2 } (\mathbf{n}_2): \langle 1, -2, 3 \rangle \\\\&\textbf{Compute the Dot Product and Magnitudes} \\&\text{The angle } \theta \text{ between the two planes is the same as the angle between their normal vectors. The cosine of} \\&\text{this angle can be found using the dot product formula:} \\&\qquad \cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1| \cdot |\mathbf{n}_2|} \\\\&\qquad \bullet \ \text{Dot Product } (\mathbf{n}_1 \cdot \mathbf{n}_2): \\&\qquad \qquad 1 \cdot 1 + 1 \cdot (-2) + 1 \cdot 3 = 1 - 2 + 3 = 2 \\\\&\qquad \bullet \ \text{Magnitude of } \mathbf{n}_1: \\&\qquad \qquad \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \\\\&\qquad \bullet \ \text{Magnitude of } \mathbf{n}_2: \\&\qquad \qquad \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \\\\&\textbf{Calculate the Angle} \\&\text{Substitute the values into the cosine formula:} \\&\qquad \cos \theta = \frac{2}{\sqrt{3} \cdot \sqrt{14}} = \frac{2}{\sqrt{42}} \\\\&\text{To find } \theta, \text{ take the inverse cosine:} \\&\qquad \theta = \cos^{-1} \left( \frac{2}{\sqrt{42}} \right)\\&\qquad \theta = \cos^{-1} \left( \sqrt{\frac{2}{{21}}} \right)\end{aligned}$$​​