The points in which the line `(x+1)/-1=(y-12)/5=(z+7)/2` cuts the surface `11x^(2)-5y^(2)+z^(2)=0` are ….......

Answer

To find the points where the line given by the equations \((x+1)/-1=(y-12)/5=(z+7)/2\) intersects the surface defined by \(11x^{2}-5y^{2}+z^{2}=0\), we can follow these steps:

### Step 1: Parameterize the Line
The line can be parameterized using a parameter \( r \):
\[ x = -r - 1, \quad y = 5r + 12, \quad z = 2r - 7 \]

### Step 2: Substitute into the Surface Equation
Substituting the parameterized coordinates into the surface equation:
\[ 11x^2 - 5y^2 + z^2 = 0 \]
we get:
\[ 11(-r - 1)^2 - 5(5r + 12)^2 + (2r - 7)^2 = 0 \]

### Step 3: Expand Each Term
Now, we will expand each term:
1. \(11(-r - 1)^2 = 11(r^2 + 2r + 1) = 11r^2 + 22r + 11\)
2. \(-5(5r + 12)^2 = -5(25r^2 + 120r + 144) = -125r^2 - 600r - 720\)
3. \((2r - 7)^2 = 4r^2 - 28r + 49\)

### Step 4: Combine the Expanded Terms
Now, combine all the expanded terms:
\[ 11r^2 + 22r + 11 - 125r^2 - 600r - 720 + 4r^2 - 28r + 49 = 0 \]
This simplifies to:
\[ (11 - 125 + 4)r^2 + (22 - 600 - 28)r + (11 - 720 + 49) = 0 \]
\[ -110r^2 - 606r - 660 = 0 \]

### Step 5: Simplify the Equation
Dividing the entire equation by -2 to simplify:
\[ 55r^2 + 303r + 330 = 0 \]

### Step 6: Use the Quadratic Formula
Now we can use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 55, b = 303, c = 330 \):
\[ r = \frac{-303 \pm \sqrt{303^2 - 4 \cdot 55 \cdot 330}}{2 \cdot 55} \]
Calculating the discriminant:
\[ 303^2 = 91809, \quad 4 \cdot 55 \cdot 330 = 72600 \]
\[ b^2 - 4ac = 91809 - 72600 = 19209 \]
Now calculate \( r \):
\[ r = \frac{-303 \pm \sqrt{19209}}{110} \]

### Step 7: Calculate the Roots
Finding the square root:
\[ \sqrt{19209} = 138.66 \quad (\text{approx.}) \]
Thus:
\[ r_1 = \frac{-303 + 138.66}{110} \quad \text{and} \quad r_2 = \frac{-303 - 138.66}{110} \]
Calculating these gives:
\[ r_1 \approx -1.49, \quad r_2 \approx -4.01 \]

### Step 8: Find the Points
Substituting \( r_1 \) and \( r_2 \) back into the parameterization:
1. For \( r_1 \):
\[ x_1 = -(-1.49) - 1 \approx 0.49, \quad y_1 = 5(-1.49) + 12 \approx 5.55, \quad z_1 = 2(-1.49) - 7 \approx -10.98 \]
2. For \( r_2 \):
\[ x_2 = -(-4.01) - 1 \approx 3.01, \quad y_2 = 5(-4.01) + 12 \approx -8.05, \quad z_2 = 2(-4.01) - 7 \approx -15.02 \]

### Final Points
Thus, the points where the line intersects the surface are approximately:
\[ (0.49, 5.55, -10.98) \quad \text{and} \quad (3.01, -8.05, -15.02) \]