To find the perpendicular distance from a point to a line in 3D space, we can use the vector formula \(d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}\), where \(A\) is a point on the line, \(P\) is the given external point, and \(\vec{b}\) is the direction vector of the line.

1. Identify the key points and vectors from the problem statement:
The given external point is \(P(1, 2, 3)\).
The equation of the line is \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\).
- From the numerators, we can identify a point on the line: \(A(6, 7, 7)\).
- From the denominators, we can identify the direction vector of the line: \(\vec{b} = \langle 3, 2, -2 \rangle\).
- Next, we form the vector that connects the point on the line (A) to the external point (P):
\[\vec{AP} = P - A = \langle 1-6, 2-7, 3-7 \rangle = \langle -5, -5, -4 \rangle\]
2. Calculate the cross product \(\vec{AP} \times \vec{b}\):
The cross product is found using the determinant of a matrix:
\[\vec{AP} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & -5 & -4 \\ 3 & 2 & -2 \end{vmatrix}\]\[= \mathbf{i}((-5)(-2) - (-4)(2)) - \mathbf{j}((-5)(-2) - (-4)(3)) + \mathbf{k}((-5)(2) - (-5)(3))\]\[= \mathbf{i}(10+8) - \mathbf{j}(10+12) + \mathbf{k}(-10+15)\]\[= 18\mathbf{i} - 22\mathbf{j} + 5\mathbf{k} = \langle 18, -22, 5 \rangle\]
3. Calculate the magnitudes of the required vectors:
First, the magnitude of the cross product vector:
\[|\vec{AP} \times \vec{b}| = \sqrt{18^2 + (-22)^2 + 5^2} = \sqrt{324 + 484 + 25} = \sqrt{833} = \sqrt{49 \times 17} = 7\sqrt{17}\]
Next, the magnitude of the line's direction vector:
\[|\vec{b}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}\]
4. Calculate the perpendicular distance:
Finally, we substitute the magnitudes into the distance formula:
\[d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|} = \frac{7\sqrt{17}}{\sqrt{17}} = 7\]
Result:
The perpendicular distance of the point P(1,2,3) from the given line is 7.