Let → a , → b and → c be three non-zero non-coplanar vectors. Let...

Given,

$\overset{\to}{a}, \overset{\to}{b} and$\overset{\to}{c}$be three non-zero non-coplanar vectors,</p><p>Also the position vectors of four points$A, B, C and D$be$\overset{\to}{a} - \overset{\to}{b} + \overset{\to}{c}, \lambda\overset{\to}{a} - 3\overset{\to}{b} + 4\overset{\to}{c}$,$- \overset{\to}{a} + 2\overset{\to}{b} - 3\overset{\to}{c}$and$2\overset{\to}{a} - 4\overset{\to}{b} + 6\overset{\to}{c}$respectively,</p><p>So,$\overset{\to}{AB} = ( \lambda - 1 ) \overset{\to}{a} - 2\overset{\to}{b} + 3\overset{\to}{c}$</p><p>$\overset{\to}{AC} = - 2\overset{\to}{a} + 3\overset{\to}{b} - 4\overset{\to}{c}$</p><p>$\overset{\to}{AD} = \overset{\to}{a} - 3\overset{\to}{b} + 5\overset{\to}{c}</p><p>Now using the condition of coplanar we get,</p><p>\left|\begin{cases}\lambda - 1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5\end{cases}\right| = 0

$⇒ ( \lambda - 1 ) ( 15 - 12 ) + 2 ( - 10 + 4 ) + 3 ( 6 - 3 ) = 0$

$⇒ ( \lambda - 1 ) = 1 ⇒ \lambda = 2$