The equation of the straight line through the origin and parallel to the line `(b+c)x+(c+a)y+(a+b)z=k=(b-c)x+(c-a)y+(a-b)z` are

Options

A. `(x)/(b^2-c^2)=(y)/(c^2-a^2)=(z)/(a^2-b^2)`
B. `(x)/(b)=(y)/(b)=(z)/(a)`
C. `(x)/(a^2-bc)=(y)/(b^2-ca)=(z)/(c^2-ab)`
D. None of these

Answer

To find the equation of the straight line through the origin and parallel to the given line, we will follow these steps:

### Step 1: Identify the equations of the planes
The given equations are:
1. \((b+c)x + (c+a)y + (a+b)z = k\)
2. \((b-c)x + (c-a)y + (a-b)z = k\)

These equations represent two planes. The line we are looking for is the intersection of these two planes.

### Step 2: Write the equations of the planes in standard form
We can rewrite the equations of the planes as:
1. \((b+c)x + (c+a)y + (a+b)z - k = 0\) (Plane 1)
2. \((b-c)x + (c-a)y + (a-b)z - k = 0\) (Plane 2)

### Step 3: Find the direction ratios of the line of intersection
The direction ratios of the line of intersection of two planes can be found using the coefficients of \(x\), \(y\), and \(z\) from the two plane equations.

Let the direction ratios of the line be given by:
\[ \text{Direction Ratios} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b+c & c+a & a+b \\ b-c & c-a & a-b \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant, we expand it:
\[ = \hat{i} \begin{vmatrix} c+a & a+b \\ c-a & a-b \end{vmatrix} - \hat{j} \begin{vmatrix} b+c & a+b \\ b-c & a-b \end{vmatrix} + \hat{k} \begin{vmatrix} b+c & c+a \\ b-c & c-a \end{vmatrix} \]

Calculating each of these 2x2 determinants gives us:
1. For \(\hat{i}\):
\[ (c+a)(a-b) - (a+b)(c-a) = ca - cb + a^2 - ab - ac + ab - b^2 + ac = a^2 - b^2 \]
2. For \(\hat{j}\):
\[ (b+c)(a-b) - (a+b)(b-c) = ba - b^2 + ca - cb - ab - b^2 + ac = c^2 - ab \]
3. For \(\hat{k}\):
\[ (b+c)(c-a) - (c+a)(b-c) = bc - ba + c^2 - ac - cb + ac + ab - c^2 = ab - ac \]

Thus, the direction ratios of the line are:
\[ \text{Direction Ratios} = (a^2 - b^2, c^2 - ab, ab - ac) \]

### Step 5: Write the equation of the line through the origin
Since the line passes through the origin, we can write the parametric equations of the line as:
\[ \frac{x}{a^2 - b^2} = \frac{y}{c^2 - ab} = \frac{z}{ab - ac} \]

### Final Answer
The equation of the straight line through the origin and parallel to the given line is:
\[ \frac{x}{a^2 - b^2} = \frac{y}{c^2 - ab} = \frac{z}{ab - ac} \]