Vectors

Step 1: Given Information
We are given the following vectors:
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-
- for some
Additionally, we are told that: $ \overrightarrow{ OC } \overrightarrow{ OB } \overrightarrow{ OA } \lambda \overrightarrow{ OC } \overrightarrow{ v_1 } = v_{1x} \hat{i} + v_{1y} \hat{j} + v_{1z} \hat{k} \overrightarrow{ v_2 } = v_{2x} \hat{i} + v_{2y} \hat{j} + v_{2z} \hat{k} |\overrightarrow{ OB } \times \overrightarrow{ OC }| \overrightarrow{ OB } \overrightarrow{ OC } \overrightarrow{ OB } = \hat{i} - 2 \hat{j} + 2 \hat{k} v_{1x} = 1 v_{1y} = -2 v_{1z} = 2 \overrightarrow{ OC } = \left( \frac{1 - 2\lambda}{2} \right) \hat{i} + \left( \frac{-2 - 2\lambda}{2} \right) \hat{j} + \left( \frac{2 - \lambda}{2} \right) \hat{k} v_{2x} = \frac{1 - 2\lambda}{2} v_{2y} = \frac{-2 - 2\lambda}{2} v_{2z} = \frac{2 - \lambda}{2} \lambda \overrightarrow{ OC } \overrightarrow{ OA } -\frac{3}{2} OAB \frac{9}{2} ABC \frac{9}{2} \overrightarrow{ OC } \overrightarrow{ OA } \overrightarrow{ OC } \cdot \overrightarrow{ OA } |\overrightarrow{ OA }| -\frac{3}{2} OAB \overrightarrow{ OA } \times \overrightarrow{ OB } \frac{9}{2} ABC \overrightarrow{ AB } = \overrightarrow{ OB } - \overrightarrow{ OA } \overrightarrow{ AC } = \overrightarrow{ OC } - \overrightarrow{ OA } \frac{9}{2} \overrightarrow{ OC } \overrightarrow{ OA } -\frac{3}{2} OAB \frac{9}{2} ABC \frac{9}{2}$ (TRUE)
Thus, all the statements are true.

Plane and Normal Vector Calculation

The correct option is (B).

Normal Vector of Plane Parallel and :

The normal vector of the plane parallel to lines and is:

Equation of Plane :

Distance from the Point (0, 1, -1) to the Plane:

(P) The distance of point (0, 1, -1) from is:

(Q) The distance from the point (0, 2, 0) to the plane is:

(R) The distance from the origin to the plane is:

Intersection Point and Distance:

(S) The point of intersection is (1, 1, 1):

Distance is: