Answer

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**Hint:**Here, in the question, we have been given position vectors of three points, which are collinear. And we are asked to find the relation between the variables present in their vectors. We will first understand the position vectors, meaning of three collinear points and then solve to find the desired relation.

**Complete step-by-step solution:**

Let the three given points be \[A,B\] and \[C\]. Therefore, their position vectors are given as,

\[\overrightarrow {OA} = x\hat i + y\hat j + z\hat k \\

\overrightarrow {OB} = \hat i + z\hat j \\

\overrightarrow {OC} = - \hat i - \hat j \]

Now, let us first understand the meaning of position vector and other terms.

Position Vector: Position vector is a vector that extends from the reference point of time to the particle. Generally, we take the origin of the coordinate system as the reference point.

Position vector of a point \[P\left( {x,y,z} \right)\] is given as \[\overrightarrow {OP} = x\hat i + y\hat j + z\hat k\] and its magnitude as \[\left| {\overrightarrow {OP} } \right| = \sqrt {{x^2} + {y^2} + {z^2}} \], where \[O\] is the origin.

Collinear points: If two or more points lie on the same line, then the points are said to be collinear.

Now, we know that, If \[{P_1}\left( {{x_1},{y_1},{z_1}} \right)\& {P_2}\left( {{x_2},{y_2},{z_2}} \right)\] are two points, then the vector joining these two points will be written as \[\overrightarrow {{P_1},{P_2}} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k\]. Therefore,

\[\overrightarrow {AB} = \left( {1 - x} \right)\hat i + \left( {z - y} \right)\hat j + \left( {0 - z} \right)\hat k \\

\overrightarrow {BC} = \left( { - 1 - 1} \right)\hat i + \left( { - 1 - z} \right)\hat j \]

Since, points \[A,B\] and \[C\] are collinear, we have \[\overrightarrow {AB} = \lambda \overrightarrow {BC} \], i.e.,

\[\left[ {\left( {1 - x} \right)\hat i + \left( {z - y} \right)\hat j + \left( { - z} \right)\hat k} \right] = \lambda \left[ {\left( { - 2} \right)\hat i + \left( { - 1 - z} \right)\hat j} \right]\]

Taking corresponding values equal, we get,

\[ - z = 0 \\

\Rightarrow z = 0 \],

\[1 - x = - 2\lambda \\

\Rightarrow x = 2\lambda + 1\],

\[ z - y = \lambda \left( { - 1 - z} \right) \\

\Rightarrow 0 - y = \lambda \left( { - 1} \right) \\

\Rightarrow y = \lambda \]

According to the values we have got for \[x,y,z\], the relation between them is \[x - 2y = 1,z = 0\].

**Hence option A). \[x - 2y = 1,z = 0\] is the correct option.**

**Note:**We had to find out the relation between two variables, that’s why we used this \[\overrightarrow {AB} = \lambda \overrightarrow {BC} \]. In case we are given three position vectors and we have to find one common variable between them, we can use the fact that the scalar triple product of all three vectors is zero if the three points given are collinear points.

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