Consider the equation,
\(\frac{x + 2}3 = \frac{2y + 3}4 = \frac{3z + 4}5 = \lambda\)
Therefore, any point on this line is of the form,
\(\left[3\lambda -2, \frac{4\lambda -3}2, \frac{5\lambda - 4}3\right]\)
Now, the line from the point (−2,3,−4) is \(3\lambda. \frac{(4\lambda - 9)}{2}, \frac{5\lambda + 8}3\)
The equation of a plane is 4x + 12y − 3z + 1 = 0
Thus, Direction ratio of normal is 4, 12, −3
Therefore,
\(4.3\lambda + 12 \left[\frac{(4\lambda - 9)}2\right] - 3\)
\(12 \lambda + 24\lambda - 54 - 5\lambda - 8 = 0\)
\(31\lambda = 62\)
\(\lambda = 2\)
Hence, the required coordinates is (4, \(\frac 52\), 2).
\(\therefore\) Distance = \(\sqrt{(4 + 2)^2 +(\frac 52 +3)^2 + (2 +4)^2}\)
\(= \sqrt{36 + 36 + \frac 14}\)
\(= \sqrt{\frac{289}4}\)
\(= \frac{17}2\)
Hence, the distance between coordinates (4, \(\frac 52\), 2) and (2, 3, −4) is \(\frac{17}2\) unit.
