VC2.10

$a\cdot b\times c=\begin{vmatrix} t & -3 & 2t \\ 1 & -2 & 2 \\3 & t & -1 \end{vmatrix}\,$

Performing $C_2\longrightarrow C_2+2C_1,C_3=C_3-2C_1\,$ we have

$\begin{vmatrix} t & 2t-3 & 0 \\ 1 & 0 & 0 \\ 3 & 6+t & -7 \end{vmatrix}\,$

$-7[0-(2t-3)]=7(2t-3)\,$

Hence $\int_1^2 a\cdot (b\times c)\,dt=7\int_1^2 (2t-3)\,dt=7[t^2-3t]_1^2\,$

= $7[(4-6)-(1-3)]=0\,$

Now $b\times c=\begin{vmatrix} i & j & k \\ 1 & -2 & 2 \\ 3 & t & 1 \end{vmatrix}\,$

= $-(2+2t)i+5j+(t+6)k\,$

Now $a\times (b\times c)=\begin{vmatrix} i & j & k \\ t & -3 & 2t \\ -(2+2t) & 5 & t+6 \end{vmatrix}\,$

= $-[(13t+18)i+5(t^2+2t)j+(t+6)k]\,$

Therefore, $\int_1^2 [a\cdot (b\times c)+a\times(b\times c)]\,dt=-\int_1^2 [(13t+18)i+5(t^2+2t)j+(t+6)k]\,dt\,$

= $-i\int_1^2 (13t+18)\,dt-5j\int_1^2 (t^2+2t)\,dt-k\int_1^2 (t+6)\,dt\,$

= $-i[\frac{13}{2}t^2+18t]_1^2-5j[\frac{1}{3}t^3+t^2]_1^2-k[\frac{1}{2}t^2+6t]_1^2\,$

= $-\frac{55}{2}i-\frac{80}{3}j-\frac{15}{2}k\,$ on simplification.