Let Q be the foot of perpendicular from the origin to the plane $$4x - 3y + z + 13 = 0$$ and R be a point (-1, -6) on the plane. Then length QR is :

To solve this problem, we need to find the length QR, where Q is the foot of the perpendicular from the origin to the plane $$4x - 3y + z + 13 = 0$$, and R is a point given as (-1, -6) on the plane. First, we must clarify the coordinates of R. Since the plane is in three dimensions and only two coordinates are provided, we need to determine the missing coordinate. The problem states "a point (-1, -6)", but it does not specify which coordinates these represent. Given that the plane equation involves x, y, and z, we must find the third coordinate such that the point lies on the plane.

Assume that the given coordinates are x and z, so R is (-1, y, -6). Substitute these into the plane equation to find y:

$$4(-1) - 3y + (-6) + 13 = 0$$

Simplify step by step:

$$-4 - 3y - 6 + 13 = 0$$

$$(-4 - 6 + 13) - 3y = 0$$

$$3 - 3y = 0$$

$$-3y = -3$$

$$y = 1$$

Thus, the coordinates of R are (-1, 1, -6). To verify, substitute back into the plane equation:

$$4(-1) - 3(1) + (-6) + 13 = -4 - 3 - 6 + 13 = -13 + 13 = 0$$

The point satisfies the plane equation, so R is (-1, 1, -6).

Next, find Q, the foot of the perpendicular from the origin O(0, 0, 0) to the plane. The normal vector to the plane $$4x - 3y + z + 13 = 0$$ is (4, -3, 1). The line from the origin along this normal has parametric equations:

$$x = 4t$$

$$y = -3t$$

$$z = t$$

Substitute these into the plane equation to find the intersection point:

$$4(4t) - 3(-3t) + t + 13 = 0$$

$$16t + 9t + t + 13 = 0$$

$$26t + 13 = 0$$

$$26t = -13$$

$$t = -\frac{13}{26} = -\frac{1}{2}$$

Now, find the coordinates of Q using t:

$$x = 4 \times -\frac{1}{2} = -2$$

$$y = -3 \times -\frac{1}{2} = \frac{3}{2}$$

$$z = -\frac{1}{2}$$

So, Q is $$\left(-2, \frac{3}{2}, -\frac{1}{2}\right)$$.

Now, compute the distance QR between Q(-2, $$\frac{3}{2}$$, -$$\frac{1}{2}$$) and R(-1, 1, -6). Use the distance formula:

$$QR = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 }$$

Substitute the coordinates:

$$QR = \sqrt{ \left(-1 - (-2)\right)^2 + \left(1 - \frac{3}{2}\right)^2 + \left(-6 - \left(-\frac{1}{2}\right)\right)^2 }$$

Simplify each difference:

$$-1 + 2 = 1 \quad \rightarrow \quad (1)^2 = 1$$

$$1 - \frac{3}{2} = -\frac{1}{2} \quad \rightarrow \quad \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$$

$$-6 + \frac{1}{2} = -\frac{12}{2} + \frac{1}{2} = -\frac{11}{2} \quad \rightarrow \quad \left(-\frac{11}{2}\right)^2 = \frac{121}{4}$$

Sum the squares:

$$1 + \frac{1}{4} + \frac{121}{4} = \frac{4}{4} + \frac{1}{4} + \frac{121}{4} = \frac{4 + 1 + 121}{4} = \frac{126}{4} = \frac{63}{2}$$

Now, take the square root:

$$QR = \sqrt{\frac{63}{2}} = \sqrt{\frac{63}{2}}$$

Simplify the expression under the square root:

$$\sqrt{\frac{63}{2}} = \sqrt{\frac{9 \times 7}{2}} = \sqrt{9} \times \sqrt{\frac{7}{2}} = 3 \sqrt{\frac{7}{2}}$$

This matches option C.

Hence, the correct answer is Option C.