Find the distance from the point \( (3,-6,-3) \) to the line \( \vec{a}(t)=\langle-2,-5,-4\rangle t+\langle 5,6,1\rangle \)

To find the distance from the point \( (3, -6, -3) \) to the line given by \( \vec{a}(t) = \langle -2, -5, -4 \rangle t + \langle 5, 6, 1 \rangle \), we start by identifying a point on the line and the direction vector of the line. 1. The direction vector \( \vec{d} \) of the line can be extracted from the expression for \( \vec{a}(t) \): \[ \vec{d} = \langle -2, -5, -4 \rangle \] 2. A point on the line can be obtained by evaluating \( \vec{a}(t) \) at \( t = 0 \): \[ \vec{r}_0 = \vec{a}(0) = \langle 5, 6, 1 \rangle \] 3. Now, we can find the vector from the point on the line \( \vec{r}_0 \) to the point \( P(3, -6, -3) \): \[ \vec{r}_0 - \vec{P} = \langle 5, 6, 1 \rangle - \langle 3, -6, -3 \rangle = \langle 2, 12, 4 \rangle \] 4. The distance \( D \) from the point to the line can be found using the formula: \[ D = \frac{|\vec{b} \cdot \vec{d}|}{|\vec{d}|} \] where \( \vec{b} = \vec{r}_0 - \vec{P} \) and \( \cdot \) denotes the dot product. 5. First, let's compute the dot product \( \vec{b} \cdot \vec{d} \): \[ \vec{b} \cdot \vec{d} = \langle 2, 12, 4 \rangle \cdot \langle -2, -5, -4 \rangle = (2)(-2) + (12)(-5) + (4)(-4) = -4 - 60 - 16 = -80 \] Therefore, \( |\vec{b} \cdot \vec{d}| = 80 \). 6. Next, we calculate the magnitude of the direction vector \( \vec{d} \): \[ |\vec{d}| = \sqrt{(-2)^2 + (-5)^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5} \] 7. Now we can substitute these values into the distance formula: \[ D = \frac{80}{3\sqrt{5}} \] 8. To simplify \( D \): \[ D = \frac{80}{3\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{80\sqrt{5}}{15} = \frac{16\sqrt{5}}{3} \] Thus, the distance from the point \( (3, -6, -3) \) to the line is: \[ \boxed{\frac{16\sqrt{5}}{3}} \]