their corresponding components must be proportional.

So,

$$ (q,-2,1)=k(4,-4,2) $$

Comparing the second and third components,

$$ -2=-4k \Rightarrow k=\frac{1}{2} $$

and

$$ 1=2k \Rightarrow k=\frac{1}{2} $$

Both give the same value of $k$, so it is consistent.

Now comparing the first component,

$$ q=4k=4\left(\frac{1}{2}\right)=2 $$

So,

$$ q=2 $$

Now the first line becomes

$$ \vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda(2\hat{i}-2\hat{j}+\hat{k}) $$

In coordinate form:

$$ x=1+2\lambda,\quad y=1-2\lambda,\quad z=-1+\lambda $$

The second line is

$$ \vec{r}=p\hat{i}-3\hat{j}+2\hat{k}+\mu(\hat{i}-2\hat{j}+2\hat{k}) $$

So,

$$ x=p+\mu,\quad y=-3-2\mu,\quad z=2+2\mu $$

Since the lines intersect, there must be common values of $\lambda$ and $\mu$ for the point of intersection.

Equating the $y$-coordinates:

$$ 1-2\lambda=-3-2\mu $$

$$ \lambda-\mu=2 $$

Equating the $z$-coordinates:

$$ -1+\lambda=2+2\mu $$

$$ \lambda-2\mu=3 $$

Now solve:

$$ \lambda-\mu=2 $$

$$ \lambda-2\mu=3 $$

Subtracting the first equation from the second:

$$ -\mu=1 $$

$$ \mu=-1 $$

Then,

$$ \lambda-\mu=2 $$

$$ \lambda+1=2 $$

$$ \lambda=1 $$

Now equate the $x$-coordinates:

$$ 1+2\lambda=p+\mu $$

Substitute $\lambda=1$ and $\mu=-1$:

$$ 1+2(1)=p+(-1) $$

$$ 3=p-1 $$

$$ p=4 $$

Thus,

$$ p=4,\quad q=2 $$

So the correct option is:

$$ \boxed{\text{Option C } \ p=4,\ q=2} $$

Three Dimensional Geometry

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