A line with positive direction cosines passes through the point $$P(2, -1, 2)$$ and makes equal angles with the coordinate axes. If the line meets the plane $$2x + y + z = 9$$ at point $$Q$$, then the length $$PQ$$ equals

The line passes through point $$P(2,-1,2)$$ and makes equal angles with the three coordinate axes.

If the direction cosines of the line are $$(l,m,n)$$, “equal angles’’ means $$l=m=n$$.
Using $$l^{2}+m^{2}+n^{2}=1$$ for direction cosines, we get $$3l^{2}=1 \;\Rightarrow\; l=m=n=\dfrac{1}{\sqrt{3}}$$.
Hence the direction ratios are proportional to $$(1,1,1)$$ (all positive as required).

So the symmetric form of the line through $$P(2,-1,2)$$ is
$$\dfrac{x-2}{1}=\dfrac{y+1}{1}=\dfrac{z-2}{1}=t$$,

or in parametric form
$$x=2+t,\; y=-1+t,\; z=2+t\qquad -(1)$$

The point $$Q$$ is where this line meets the plane $$2x+y+z=9$$.
Substituting from $$(1)$$ into the plane equation:

$$2(2+t)+(-1+t)+(2+t)=9$$
$$4+2t-1+t+2+t=9$$
$$5+4t=9$$
$$4t=4\;\Rightarrow\; t=1$$

Putting $$t=1$$ back in $$(1)$$, the coordinates of $$Q$$ are
$$Q(\,2+1,\;-1+1,\;2+1\,)=(3,0,3)$$.

The length $$PQ$$ is the distance between $$P(2,-1,2)$$ and $$Q(3,0,3)$$:

$$PQ=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \;=\; \sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$$.

Therefore, the required length is $$\sqrt{3}$$.

Option C which is: $$\sqrt{3}$$