Let `A(1, 2, 3), B(0, 0, 1) and C(-1, 1, 1)` are the vertices of `triangleABC`.
Q. The equation of altitude through B to side AC is

Options

A. `r=k+t(7hat(i)-10hat(j)+2hat(k))`
B. `r=k+t(-7hat(i)+10hat(j)+2hat(k))`
C. `r=k+t(7hat(i)-10hat(j)-2hat(k))`
D. `r=k+t(7hat(i)+10hat(j)+2hat(k))`

Answer

To find the equation of the altitude through point B to side AC of triangle ABC with vertices A(1, 2, 3), B(0, 0, 1), and C(-1, 1, 1), we will follow these steps:

### Step 1: Find the equation of line AC
The line AC can be represented in the symmetric form. The direction ratios of line AC can be found using the coordinates of points A and C.

- **Coordinates of A**: (1, 2, 3)
- **Coordinates of C**: (-1, 1, 1)

The direction ratios of AC can be calculated as follows:
- Change in x = -1 - 1 = -2
- Change in y = 1 - 2 = -1
- Change in z = 1 - 3 = -2

Thus, the direction ratios of AC are (-2, -1, -2).

Using the symmetric form of the line equation:
\[ \frac{x - 1}{-2} = \frac{y - 2}{-1} = \frac{z - 3}{-2} \]

### Step 2: Parametrize the line AC
Let \( r \) be the parameter. We can express the coordinates of any point D on line AC in terms of \( r \):
- \( x = -2r + 1 \)
- \( y = -r + 2 \)
- \( z = -2r + 3 \)

### Step 3: Find the direction ratios of line BD
The direction ratios of line BD can be found using the coordinates of points B and D.

- **Coordinates of B**: (0, 0, 1)
- **Coordinates of D**: \((-2r + 1, -r + 2, -2r + 3)\)

The direction ratios of BD are:
- Change in x = \((-2r + 1) - 0 = -2r + 1\)
- Change in y = \((-r + 2) - 0 = -r + 2\)
- Change in z = \((-2r + 3) - 1 = -2r + 2\)

Thus, the direction ratios of BD are \((-2r + 1, -r + 2, -2r + 2)\).

### Step 4: Find the direction ratios of line AC
From Step 1, we already have the direction ratios of line AC as (-2, -1, -2).

### Step 5: Set up the perpendicularity condition
Since BD is perpendicular to AC, their dot product must equal zero:
\[ (-2)(-2r + 1) + (-1)(-r + 2) + (-2)(-2r + 2) = 0 \]

Expanding this gives:
\[ 4r - 2 + r - 2 + 4r - 4 = 0 \]
Combining like terms:
\[ 9r - 8 = 0 \]
Thus, solving for \( r \):
\[ r = \frac{8}{9} \]

### Step 6: Find the coordinates of point D
Substituting \( r = \frac{8}{9} \) back into the equations for D:
- \( x = -2\left(\frac{8}{9}\right) + 1 = -\frac{16}{9} + 1 = -\frac{16}{9} + \frac{9}{9} = -\frac{7}{9} \)
- \( y = -\left(\frac{8}{9}\right) + 2 = -\frac{8}{9} + \frac{18}{9} = \frac{10}{9} \)
- \( z = -2\left(\frac{8}{9}\right) + 3 = -\frac{16}{9} + \frac{27}{9} = \frac{11}{9} \)

Thus, the coordinates of point D are \((-7/9, 10/9, 11/9)\).

### Step 7: Write the equation of line BD
Using the coordinates of B and D, we can write the equation of line BD in symmetric form:
\[ \frac{x - 0}{-7/9} = \frac{y - 0}{10/9} = \frac{z - 1}{2/9} \]

This can be simplified to:
\[ \frac{x}{-7} = \frac{y}{10} = \frac{z - 1}{2} \]

### Final Answer
The equation of the altitude through B to side AC is:
\[ \frac{x}{-7} = \frac{y}{10} = \frac{z - 1}{2} \]