Consider a plane `prod:vecr*(2hati+hatj-hatk)=5`, a line `L_(1): vecr=(3hati-hatj+2hatk)+k(2hati-3hatj-hatk)` and a point `A(3, -4, 1)*L_(2)` is a line passing through A intersecting `L_(1)` and parallel to plane `prod`.
Q. Line `L_(1)` intersects plane `prod` at Q and xy-plane at R the volume of tetrahedron OAQR is :
(where 'O' is origin)

Options

A. (a) 0
B. (b) `(14)/(3)`
C. (c) `(3)/(7)`
D. (d) `(7)/(3)`

Answer

To find the volume of the tetrahedron \( OAQR \) where \( O \) is the origin, we need to follow these steps:

### Step 1: Find the intersection point \( Q \) of line \( L_1 \) and the plane \( \pi \).

The plane \( \pi \) is given by the equation:
\[ \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) = 5 \]
This can be expressed in Cartesian form as:
\[ 2x + y - z = 5 \]

The line \( L_1 \) is given by:
\[ \vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + k(2\hat{i} - 3\hat{j} - \hat{k}) \]
In Cartesian form, this can be expressed as:
\[ x = 3 + 2k, \quad y = -1 - 3k, \quad z = 2 - k \]

Substituting these into the plane equation:
\[ 2(3 + 2k) + (-1 - 3k) - (2 - k) = 5 \]
Simplifying:
\[ 6 + 4k - 1 - 3k - 2 + k = 5 \]
\[ 4k - 3k + k + 3 = 5 \]
\[ 2k + 3 = 5 \implies 2k = 2 \implies k = 1 \]

Now substituting \( k = 1 \) back into the equations of \( L_1 \):
\[ x = 3 + 2(1) = 5, \quad y = -1 - 3(1) = -4, \quad z = 2 - 1 = 1 \]
Thus, the coordinates of point \( Q \) are \( (5, -4, 1) \).

### Step 2: Find the intersection point \( R \) of line \( L_1 \) with the XY-plane.

The XY-plane is defined by \( z = 0 \). Setting \( z = 0 \) in the equations of \( L_1 \):
\[ 0 = 2 - k \implies k = 2 \]
Now substituting \( k = 2 \) back into the equations of \( L_1 \):
\[ x = 3 + 2(2) = 7, \quad y = -1 - 3(2) = -7, \quad z = 0 \]
Thus, the coordinates of point \( R \) are \( (7, -7, 0) \).

### Step 3: Find the volume of tetrahedron \( OAQR \).

The volume \( V \) of tetrahedron formed by points \( O(0, 0, 0) \), \( A(3, -4, 1) \), \( Q(5, -4, 1) \), and \( R(7, -7, 0) \) can be calculated using the scalar triple product:
\[ V = \frac{1}{6} | \vec{OA} \cdot (\vec{OQ} \times \vec{OR}) | \]

Calculating the vectors:
\[ \vec{OA} = (3, -4, 1), \quad \vec{OQ} = (5, -4, 1), \quad \vec{OR} = (7, -7, 0) \]

### Step 4: Calculate \( \vec{OQ} \times \vec{OR} \).

Using the determinant to find the cross product:
\[ \vec{OQ} \times \vec{OR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -4 & 1 \\ 7 & -7 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}((-4)(0) - (1)(-7)) - \hat{j}((5)(0) - (1)(7)) + \hat{k}((5)(-7) - (-4)(7)) \]
\[ = \hat{i}(0 + 7) - \hat{j}(0 - 7) + \hat{k}(-35 + 28) \]
\[ = 7\hat{i} + 7\hat{j} - 7\hat{k} = 7(1, 1, -1) \]

### Step 5: Calculate \( \vec{OA} \cdot (\vec{OQ} \times \vec{OR}) \).

Now calculate the dot product:
\[ \vec{OA} \cdot (7, 7, -7) = (3)(7) + (-4)(7) + (1)(-7) \]
\[ = 21 - 28 - 7 = -14 \]

### Step 6: Calculate the volume.

Thus, the volume is:
\[ V = \frac{1}{6} | -14 | = \frac{14}{6} = \frac{7}{3} \]

### Final Answer:
The volume of tetrahedron \( OAQR \) is \( \frac{7}{3} \).

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