Question 15 (10 points) Determinethe distance between: a) the point $E(1,3,-2)$ to the line $\vec{r}=(1,3,1)+t(1,1,-5), t \in \Re$. [5] b) the planes $\Pi_{1}: 2 x+2 y-z-5=0$ and $\Pi_{2}: 4 x+4 y-2 z+17=0$. [3] c) the line $\vec{r}=(1,3,2)+t(4,2,-1), t \in \mathfrak{R}$ and the plane $y+2 z-5=0$. [2]

Step 1 ：Find the distance between point E and the line by finding the vector from E to a point on the line, projecting that vector onto the direction vector of the line, and finding the magnitude of the difference between the original vector and the projection.

Step 2 ：\(E = (1, 3, -2)\)

Step 3 ：\(\vec{r} = (1, 3, 1) + t(1, 1, -5)\)

Step 4 ：\(\text{vector to line} = (0, 0, 3)\)

Step 5 ：\(\text{projection} = (-0.5556, -0.5556, 2.7778)\)

Step 6 ：\(\boxed{\text{distance}_a \approx 0.8165}\)

Step 7 ：Find the distance between the two planes by substituting a point from one plane into the equation of the other plane and dividing by the magnitude of the normal vector.

Step 8 ：\(\Pi_1: 2x + 2y - z - 5 = 0\)

Step 9 ：\(\Pi_2: 4x + 4y - 2z + 17 = 0\)

Step 10 ：\(\boxed{\text{distance}_b \approx 1.1667}\)

Step 11 ：Find the distance between the line and the plane by finding the point on the line that is closest to the plane and finding the distance between that point and the plane.

Step 12 ：\(\vec{r}_c = (1, 3, 2) + t(4, 2, -1)\)

Step 13 ：\(y + 2z - 5 = 0\)

Step 14 ：\(t_{\text{closest}} = 2.6180\)

Step 15 ：\(\text{point closest} = (11.4721, 8.2361, -0.6180)\)

Step 16 ：\(\boxed{\text{distance}_c \approx 5.3666}\)