The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines $$\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$$ and $$\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$$ is

We have to find the length of the perpendicular drawn from the point $$P(2,1,4)$$ to the plane that contains the two given lines

$$\vec r=(\hat i+\hat j)+\lambda(\hat i+2\hat j-\hat k)$$

and

$$\vec r=(\hat i+\hat j)+\mu(-\hat i+\hat j-2\hat k).$$

First we translate each vector equation into a more convenient Cartesian form. The position vector $$\hat i+\hat j$$ corresponds to the fixed point $$A(1,1,0).$$ The direction vector of the first line is $$\vec v_1=\langle1,\,2,\,-1\rangle,$$ while the direction vector of the second line is $$\vec v_2=\langle-1,\,1,\,-2\rangle.$$ Since both lines start at the same point $$A(1,1,0),$$ they intersect there, and the required plane certainly passes through point $$A.$$

To get the equation of the plane, we need a normal vector. A normal vector $$\vec n$$ can be obtained by the cross-product of the two non-parallel direction vectors. We recall the formula for the cross-product:

$$\vec v_1\times\vec v_2= \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] v_{1x} & v_{1y} & v_{1z}\\[2pt] v_{2x} & v_{2y} & v_{2z} \end{vmatrix}.$$

Substituting $$\vec v_1=\langle1,2,-1\rangle$$ and $$\vec v_2=\langle-1,1,-2\rangle,$$ we have

$$ \vec n = \bigl(2\cdot(-2)-(-1)\cdot1\bigr)\hat i - \bigl(1\cdot(-2)-(-1)\cdot(-1)\bigr)\hat j + \bigl(1\cdot1-2\cdot(-1)\bigr)\hat k. $$

Simplifying term by term,

$$ 2\cdot(-2)-(-1)\cdot1 = -4+1=-3,\\[4pt] 1\cdot(-2)-(-1)\cdot(-1) = -2-1=-3,\;\text{and hence}\;-\bigl(-3\bigr)=+3,\\[4pt] 1\cdot1-2\cdot(-1)=1+2=3. $$

So we obtain

$$\vec n=\langle-3,3,3\rangle.$$

A normal vector may be scaled by any non-zero constant; dividing each component by $$-3$$ gives the shorter normal $$\langle1,-1,-1\rangle.$$ Both choices are equivalent, but the simpler one often shortens later arithmetic. Let us therefore use $$\vec n=\langle1,-1,-1\rangle.$$

With normal vector $$\langle1,-1,-1\rangle$$ and point $$A(1,1,0),$$ the plane equation follows from

$$\vec n\cdot\bigl\langle x-1,\;y-1,\;z-0\bigr\rangle=0.$$

Carrying out the dot product,

$$1(x-1)+(-1)(y-1)+(-1)(z)=0,$$

which expands to

$$x-1-y+1-z=0\;\;\Longrightarrow\;\;x-y-z=0.$$

Thus the required plane is

$$x-y-z=0.$$

Now we invoke the distance formula from a point to a plane. For a plane $$ax+by+cz+d=0$$ and a point $$P(x_0,y_0,z_0),$$ the perpendicular distance $$D$$ is

$$D=\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}.$$

In our case $$a=1,\;b=-1,\;c=-1,\;d=0,$$ and $$P(2,1,4).$$ Substituting each value gives

$$D=\dfrac{|1(2)+(-1)(1)+(-1)(4)+0|}{\sqrt{1^{2}+(-1)^{2}+(-1)^{2}}}.$$

Evaluating the numerator step by step,

$$1(2)=2,\quad(-1)(1)=-1,\quad(-1)(4)=-4.$$ So $$2-1-4=-3,$$ and therefore $$|2-1-4|=|-3|=3.$$

For the denominator,

$$1^{2}+(-1)^{2}+(-1)^{2}=1+1+1=3,$$ so $$\sqrt{1^{2}+(-1)^{2}+(-1)^{2}}=\sqrt{3}.$$

Combining numerator and denominator,

$$D=\dfrac{3}{\sqrt3}.$$

We rationalise if desired: multiplying numerator and denominator by $$\sqrt3$$ gives

$$D=\dfrac{3\sqrt3}{3}=\sqrt3.$$

Hence the length of the perpendicular from the point $$P(2,1,4)$$ to the plane is $$\sqrt3.$$

Therefore the required answer corresponds to Option C.

Hence, the correct answer is Option C.