The angle between the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and the plane $x+y+z=0$ is:

Solution

1. The direction vector of the line is $\vec{v}=(1,2,3)$ and the normal vector of the plane $x+y+z=0$ is $\vec{n}=(1,1,1)$.
2. The sine of the angle $\theta$ between the line and the plane is given by $\sin \theta =\frac{|\vec{v}\cdot \vec{n}|}{|\vec{v}||\vec{n}|}$.
3. Calculating the dot product and magnitudes, we get $$\begin{array}{rl}\sin \theta & =\frac{|(1)(1)+(2)(1)+(3)(1)|}{\sqrt{1^2+2^2+3^2}\sqrt{1^2+1^2+1^2}}\\ & =\frac{6}{\sqrt{14}\sqrt{3}}\\ & =\frac{6}{\sqrt{42}}\end{array}$$.
4. Simplifying the expression, $$\begin{array}{rl}\frac{6}{\sqrt{42}}& =\sqrt{\frac{36}{42}}\\ & =\sqrt{\frac{6}{7}}\end{array}$$.
5. Hence the answer is (A).