(a) If $A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 2 & 0 & -3 \\ 1 & 2 & 0\end{array}\right]$, then find $A^{-1}$ and hence solve the following system of equations:

$$\begin{array}{r} x+2 y-3 z=1 \\ 2 x-3 z=2 \\ x+2 y=3 \end{array}$$

OR

(b) Find the product of the matrices $\left[\begin{array}{ccc}1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}-6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1\end{array}\right]$ and hence solve the system of linear equations:

$$\begin{aligned} x+2 y-3 z & =4 \\ 2 x+3 y+2 z & =2 \\ 3 x-3 y-4 z & =11 \end{aligned}$$

Find the area of the region bounded by the curve $4 x^2$ $+y^2=36$ using integration.

(a) Find the co-ordinates of the foot of the perpendicular drawn from the point $(2,3,-8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$.

OR

(b). Find the shortest distance between the lines $\mathrm{L}_1 \& \mathrm{~L}_2$ given below: $\mathrm{L}_1$ : The line passing through $(2,-1,1)$ and parallel to $\frac{x}{1}=\frac{y}{1}=\frac{z}{3} \mathrm{~L}_2: \vec{r}=\hat{i}+(2 \mu+1) \hat{j}-(\mu+2) \hat{k}$.

$$\begin{aligned} &\text { Solve the following L. P. P. graphically: }\\ &\begin{array}{rlrl} \text { Maximise } & Z =60 x+40 y \\ \text { Subject to } & x+2 y \leq 12 \\ & 2 x+y \leq 12 \\ & 4 x+5 y \geq 20 \\ & x, y \geq 0 \end{array} \end{aligned}$$