Find the (perpendicular) distance from the line given by the parametric equations \[ \begin{array}{l}x(t)=-5+2 t \\ y(t)=2+7 t \\ z(t)=5+t\end{array} \] to the point \( (7,2,-10) \).

To find the perpendicular distance from the line given by the parametric equations to the point \( (7,2,-10) \), we can use the following method: 1. **Define the points and direction vector**: The parametric line can be represented as: \[ \mathbf{p}(t) = (-5 + 2t, 2 + 7t, 5 + t). \] The point on the line when \( t = 0 \) is \( (-5, 2, 5) \) and the direction vector of the line, derived from the coefficients of \( t \), is: \[ \mathbf{d} = (2, 7, 1). \] 2. **Find the vector from the line to the point**: The vector \( \mathbf{v} \) from a point on the line \( (-5, 2, 5) \) to the point \( (7, 2, -10) \) is: \[ \mathbf{v} = (7 - (-5), 2 - 2, -10 - 5) = (12, 0, -15). \] 3. **Calculate the projection of vector \(\mathbf{v}\) onto direction vector \(\mathbf{d}\)**: We need to calculate the projection of \( \mathbf{v} \) onto \( \mathbf{d} \): \[ \text{proj}_{\mathbf{d}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}. \] First, calculate \( \mathbf{v} \cdot \mathbf{d} \): \[ \mathbf{v} \cdot \mathbf{d} = (12 \cdot 2) + (0 \cdot 7) + (-15 \cdot 1) = 24 - 15 = 9. \] Next, calculate \( \mathbf{d} \cdot \mathbf{d} \): \[ \mathbf{d} \cdot \mathbf{d} = 2^2 + 7^2 + 1^2 = 4 + 49 + 1 = 54. \] Now we can find the projection: \[ \text{proj}_{\mathbf{d}} \mathbf{v} = \frac{9}{54} \mathbf{d} = \frac{1}{6} (2, 7, 1) = \left( \frac{1}{3}, \frac{7}{6}, \frac{1}{6} \right). \] 4. **Calculate the component of \(\mathbf{v}\) orthogonal to \(\mathbf{d}\)**: The orthogonal component \( \mathbf{v}_{\perp} \) is given by: \[ \mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_{\mathbf{d}} \mathbf{v}. \] Thus: \[ \mathbf{v}_{\perp} = (12, 0, -15) - \left( \frac{1}{3}, \frac{7}{6}, \frac{1}{6} \right) = \left( 12 - \frac{1}{3}, 0 - \frac{7}{6}, -15 - \frac{1}{6} \right). \] This simplifies to: \[ \mathbf{v}_{\perp} = \left( \frac{36 - 1}{3}, -\frac{7}{6}, -\frac{90 + 1}{6} \right) = \left( \frac{35}{3}, -\frac{7}{6}, -\frac{91}{6} \right). \] 5. **Calculate the distance**: Now we find the magnitude of \( \mathbf{v}_{\perp} \): \[ \|\mathbf{v}_{\perp}\| = \sqrt{ \left( \frac{35}{3} \right)^2 + \left( -\frac{7}{6} \right)^2 + \left( -\frac{91}{6} \right)^2 }. \] This calculates to: \[ \|\mathbf{v}_{\perp}\| = \sqrt{ \frac{1225}{9} + \frac{49}{36} + \frac{828