Let P be a plane containing the line x - 1 3 = y + 6 4 = z + 5 2 and parallel to the line x - 1 4 = y - 2 - 3 = z + 5 7. If the point (1, -1, ) lies on the plane P, then the value of |5| is equal to ____________.

Equation of required plane is \\ vmatrix x - 1 & y + 6 & z + 5 \\ 3 & 4 & 2 \\ 4 & -3 & 7 vmatrix = 0 Since, (1, -1, ) lies on

Let P be a plane containing the line {x - 1} \ 3} = {{y + 6} \ 4} = {{z + 5} \ 2} and parallel to the line: (Mathematics)

Equation of required plane is $\\ \begin{vmatrix} x - 1 & y + 6 & z + 5 \\ 3 & 4 & 2 \\ 4 & -3 & 7 \end{vmatrix} = 0$

Since, (1, $-$ 1, $\alpha$ ) lies on it,

So, replace x by 1, y by ( $-1$ ) and z and $\alpha$ .

$\begin{vmatrix} 0 & 5 & \alpha + 5 \\ 3 & 4 & 2 \\ 4 & -3 & 7 \end{vmatrix} = 0$

$\Rightarrow 5\alpha + 38 = 0 \Rightarrow 5\alpha = - 38$

$\therefore \left| {5\alpha } \right| = \left| { - 38} \right| = 38$