Answer:
Option C
Explanation:
(P) Given ,[a b c]=2
$\therefore $ Volume of
2 (a x b),3(b x c), (c x a)
$\Rightarrow$ 6[axb bxc cxa]
$\Rightarrow$ 6[a b c]2
$\Rightarrow$ 6 x 4=24
(Q) [a b c] =5 , given
$\therefore$ Volume of
3(a+b),(b+c),2(c+a)
$\Rightarrow$ 6[a+b b+c c+a]
$\Rightarrow$ 6 x2 [a b c]
$\Rightarrow$ 12 x 5=60
(R) $\frac{1}{2}|a\times b|=20$ , given
$\therefore$ $\triangle_{1}=\frac{1}{2}|(2a+3b)\times(a-b)|$
$=\frac{1}{2}|2a\times a-2a \times b+3b \times a-3b\times b|$
$=\frac{1}{2}|2b\times a+3b \times a|=\frac{5}{2}|a\times b|$
=5 x20=100
(S) Given, |a x b|=30
$\therefore$ |(a+b)x a|=|axb + bxa|
= |b x a|
= |a x b|=30
