(a) Find the length of the perpendicular from the point (1,2,3) to the line :
`(x - 6)/(3) = (y -7)/(2) = (z - 7)/(-2)` .
(b) Find the perpendicular distance from the point (1,2,3) to the line :
`vec(r) = 6 hati +7 hatj+ 7 hatk + lambda (3 hati + 2 hatj - 2 hatk ). `

Answer

To solve the problem, we will break it down into two parts as given in the question.

### Part (a): Find the length of the perpendicular from the point (1,2,3) to the line given by:
\[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2} \]

**Step 1: Identify a point and direction vector on the line.**
- The line can be expressed in parametric form. Let \( t \) be the parameter:
- \( x = 3t + 6 \)
- \( y = 2t + 7 \)
- \( z = -2t + 7 \)

From this, we can identify:
- A point on the line \( A(6, 7, 7) \) when \( t = 0 \).
- The direction vector \( \vec{b} = (3, 2, -2) \).

**Step 2: Find the coordinates of a general point \( B \) on the line.**
- For any value of \( t \), the coordinates of point \( B \) are:
\[ B(3t + 6, 2t + 7, -2t + 7) \]

**Step 3: Find the vector \( \vec{BP} \) from point \( P(1, 2, 3) \) to point \( B \).**
- The vector \( \vec{BP} \) is given by:
\[ \vec{BP} = (3t + 6 - 1, 2t + 7 - 2, -2t + 7 - 3) = (3t + 5, 2t + 5, -2t + 4) \]

**Step 4: Set up the equation for the perpendicular distance.**
- The direction vector of the line \( \vec{b} = (3, 2, -2) \).
- For \( \vec{BP} \) to be perpendicular to \( \vec{b} \), their dot product must be zero:
\[ \vec{BP} \cdot \vec{b} = 0 \]
\[ (3t + 5) \cdot 3 + (2t + 5) \cdot 2 + (-2t + 4) \cdot (-2) = 0 \]

**Step 5: Solve the dot product equation.**
- Expanding the equation:
\[ 9t + 15 + 4t + 10 + 4t - 8 = 0 \]
\[ 17t + 17 = 0 \]
\[ t = -1 \]

**Step 6: Find the coordinates of point \( B \) when \( t = -1 \).**
- Substituting \( t = -1 \):
\[ B(3(-1) + 6, 2(-1) + 7, -2(-1) + 7) = (3, 5, 9) \]

**Step 7: Calculate the distance \( D \) between points \( P(1, 2, 3) \) and \( B(3, 5, 9) \).**
- The distance formula is:
\[ D = \sqrt{(3 - 1)^2 + (5 - 2)^2 + (9 - 3)^2} \]
\[ D = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Part (b): Find the perpendicular distance from the point (1,2,3) to the line given by:
\[ \vec{r} = 6 \hat{i} + 7 \hat{j} + 7 \hat{k} + \lambda (3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \]

**Step 1: Identify a point and direction vector on the line.**
- The fixed point on the line is \( A(6, 7, 7) \).
- The direction vector is \( \vec{b} = (3, 2, -2) \).

**Step 2: Since the line is the same as in part (a), the distance will also be the same.**
- The point \( P(1, 2, 3) \) and the line are unchanged.

**Step 3: Conclude the distance.**
- Therefore, the perpendicular distance from point \( P(1, 2, 3) \) to this line is also \( 7 \).

### Final Answers:
(a) The length of the perpendicular from the point (1,2,3) to the line is **7**.
(b) The perpendicular distance from the point (1,2,3) to the line is also **7**.