Answer

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**Hint:**This question belongs to the topic of calculus. In this question, we will first find the direction vector from the three equations. These direction vectors will be used for finding the new vector equations. After that we will use the same direction vector and the point \[\left( 0,14,-10 \right)\] to find the parametric equation. After using the parametric equation, we will find the vector equation of a line.

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

Let us solve this question.

In this question, we have to find the vector equation and parametric equations for the line. It is given in the question that the line is passing through the point \[\left( 0,14,-10 \right)\] and is parallel to the line having parametric equations as \[x=-1+2t\], \[y=6-3t\], \[z=3+9t\].

So, let us first find out the direction vector from the above parametric equations.

The direction vectors from the parametric equations will be \[\left( 2,-3,9 \right)\].

As we have to find the line which is parallel to the line having parametric equations \[x=-1+2t\], \[y=6-3t\], and \[z=3+9t\]. Then, both of them will have same direction vectors.

So, the parametric equations for the line we have to find will be \[x={{x}_{0}}+2t\], \[y={{y}_{0}}-3t\], and \[z={{z}_{0}}+9t\], where the point \[\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\] is passing through the same line.

It is said in the question that the line is passing through the point \[\left( 0,14,-10 \right)\]

So, the parametric equations of the line which is passing through the point \[\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\] and is parallel to the line (whose parametric equations are given as \[x=-1+2t\], \[y=6-3t\], and \[z=3+9t\] in the question) will be \[x=0+2t\], \[y=14-3t\], and \[z=-10+9t\].

We have asked to find the vector equation also.

As we know that the 3 dimensional vectors are always in the form of \[x\overset{\hat{\ }}{\mathop{i}}\,+y\overset{\hat{\ }}{\mathop{j}}\,+z\overset{\hat{\ }}{\mathop{k}}\,\]

So, we can write the above vector by putting the value of x and y as

\[\left( 0+2t \right)\overset{\hat{\ }}{\mathop{i}}\,+\left( 14-3t \right)\overset{\hat{\ }}{\mathop{j}}\,+\left( -10+9t \right)\overset{\hat{\ }}{\mathop{k}}\,\]

Which is also can be written as

\[\left( 0\overset{\hat{\ }}{\mathop{i}}\,+14\overset{\hat{\ }}{\mathop{j}}\,-10\overset{\hat{\ }}{\mathop{k}}\, \right)+t\left( 2\overset{\hat{\ }}{\mathop{i}}\,-3\overset{\hat{\ }}{\mathop{j}}\,+9\overset{\hat{\ }}{\mathop{k}}\, \right)\]

**Note:**As this question belongs to the topic of calculus and vectors, so we should have better knowledge in those topics. Whenever it is asked to find a line which is parallel to the line whose parametric equations are: \[x={{x}_{0}}+2t,y={{y}_{0}}-3t,z={{z}_{0}}+9t\]. It is also said that the line is passing through the point \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\]. Then, the parametric equation of the new line will be \[x={{x}_{1}}+2t,y={{y}_{1}}-3t,z={{z}_{1}}+9t\].

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