Let P be a plane containing the line (x-1)/3 = (y+6)/4 = (z+5)/2 and parallel to the line .

Question

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

Answer 1

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

Now (1, –1, α) lies on it so