Q.(a) Find the vector and Cartesian equation of the plane passing through the point $(0, 1, -5)$ and parallel to the straight lines $\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+s(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r}=(\hat{i}-3\hat{j}+5\hat{k})+t(\hat{i}+\hat{j}-\hat{k})$ OR

(a) finds the plane's normal as the cross product of the two given lines' direction vectors and writes the plane through the given point; (b) evaluates the definite integral using the standard King's-rule trick for $\int_{-a}^a f(x)/(1+k^x)dx$ with even $f$.

(a) Plane parallel to two lines, through $(0,1,-5)$

1. The two lines have direction vectors $\vec d_1=2\hat i+3\hat j+6\hat k$ and $\vec d_2=\hat i+\hat j-\hat k$. A plane parallel to both lines has normal $\vec n=\vec d_1\times\vec d_2$.
2. $\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&6\\1&1&-1\end{vmatrix}=\hat i(3(-1)-6(1))-\hat j(2(-1)-6(1))+\hat k(2(1)-3(1))=\hat i(-9)-\hat j(-8)+\hat k(-1)=-9\hat i+8\hat j-\hat k$.
3. Vector equation: $\vec r\cdot\vec n=\vec a\cdot\vec n$, where $\vec a=0\hat i+\hat j-5\hat k$ is the given point's position vector. $\vec a\cdot\vec n=0(-9)+1(8)+(-5)(-1)=8+5=13$.
4. So $\vec r\cdot(-9\hat i+8\hat j-\hat k)=13$.
5. Cartesian form (with $\vec r=x\hat i+y\hat j+z\hat k$): $-9x+8y-z=13$, i.e. $9x-8y+z+13=0$. Check at $(0,1,-5)$: $0-8-5+13=0$. Verified.

(b) Evaluate $\int_{-\pi}^{\pi}\dfrac{\cos^2x}{1+a^x}dx$ …