The perpendicular distance, of the line $$\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$$ from the point P(2,−10, 1), is :

The perpendicular distance from a point to a line in 3D space is given by the formula:

$$d = \frac{|\overrightarrow{QP} \times \vec{d}|}{|\vec{d}|}$$

where:

• Q is a point on the line,
• $$\vec{d}$$ is the direction vector of the line,
• P is the given point.

First, identify a point on the line. The line is given in symmetric form:

$$\frac{x-1}{2} = \frac{y+2}{-1} = \frac{z+3}{2}$$

Set the parameter $$\lambda = 0$$ to find a point on the line:

When $$\lambda = 0$$:

$$x - 1 = 0 \implies x = 1$$

$$y + 2 = 0 \implies y = -2$$

$$z + 3 = 0 \implies z = -3$$

Thus, point Q is $$(1, -2, -3)$$.

The direction vector $$\vec{d}$$ is given by the denominators: $$(2, -1, 2)$$, so $$\vec{d} = 2\hat{i} - \hat{j} + 2\hat{k}$$.

The given point is P$$(2, -10, 1)$$. The vector $$\overrightarrow{QP}$$ is from Q to P:

$$\overrightarrow{QP} = P - Q = (2 - 1, -10 - (-2), 1 - (-3)) = (1, -8, 4)$$

So, $$\overrightarrow{QP} = \hat{i} - 8\hat{j} + 4\hat{k}$$.

Now, compute the cross product $$\overrightarrow{QP} \times \vec{d}$$:

$$\overrightarrow{QP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \\ \end{vmatrix}$$

Expand the determinant:

$$= \hat{i} \left[ (-8)(2) - (4)(-1) \right] - \hat{j} \left[ (1)(2) - (4)(2) \right] + \hat{k} \left[ (1)(-1) - (-8)(2) \right]$$

Calculate each component:

$$\hat{i}$$ component: $$(-16) - (-4) = -16 + 4 = -12$$

$$\hat{j}$$ component: $$-\left[ 2 - 8 \right] = -[-6] = 6$$ (since the $$\hat{j}$$ term has a negative sign in the expansion)

$$\hat{k}$$ component: $$[-1] - [-16] = -1 + 16 = 15$$

Thus, $$\overrightarrow{QP} \times \vec{d} = -12\hat{i} + 6\hat{j} + 15\hat{k}$$.

Now, find the magnitude:

$$|\overrightarrow{QP} \times \vec{d}| = \sqrt{(-12)^2 + (6)^2 + (15)^2} = \sqrt{144 + 36 + 225} = \sqrt{405}$$

Simplify $$\sqrt{405}$$:

$$\sqrt{405} = \sqrt{81 \times 5} = 9\sqrt{5}$$

Next, find the magnitude of $$\vec{d}$$:

$$|\vec{d}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$

Finally, the perpendicular distance is:

$$d = \frac{|\overrightarrow{QP} \times \vec{d}|}{|\vec{d}|} = \frac{9\sqrt{5}}{3} = 3\sqrt{5}$$

The options are:

A. 6

B. $$5\sqrt{2}$$

C. $$4\sqrt{3}$$

D. $$3\sqrt{5}$$

The distance $$3\sqrt{5}$$ matches option D.