Let the line L pass through the point ( 0, 1, 2 ) , intersect the...

Given,

The line $L$ pass through the point $( 0, 1, 2 ) ,$ intersect the line $\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}$ and be parallel to the plane $2x + y - 3z = 4$,

Now plotting the diagram of above data we have,

Now let the direction of line which passes through $\left(0, 1, 2\right)$ be \left(a, b, c\right)$</p><p>Now the line$L \text{ and } {L}_{1}are coplanar,</p><p>So coplanar condition we get,</p><p>\left|\begin{cases}a & b & c \\ 2 - 1 & 3 - 2 & 4 - 3 \\ 2 & 3 & 4\end{cases}\right| = 0

⇒ \left|\begin{cases}a & b & c \\ 1 & 1 & 1 \\ 2 & 3 & 4\end{cases}\right| = 0

$⇒ a - 2b + c = 0 . . . . . . . \left(1\right)$

And the line $L$ is perpendicular to the normal vector of the plane,

So by perpendicular condition we get,
$2a + b - 3c = 0 . . . . . . \left(2\right)</p><p>So, from equation \left(1\right) \text{\text{ and } }$ \left(2\right)$we get,</p><p>$a = b = c$</p><p>So, equation of line$L$will be,</p><p>$⇒ L = \dfrac{x}{1} = \dfrac{y - 1}{1} = \dfrac{z - 2}{1} =\lambda$</p><p>So any point on$L$can be taken as A\left(\lambda, 1 + \lambda, 2 + \lambda\right)$,

Now $\overset{\to}{AP}$ will be perpendicular to the direction ratio of the line $L$,

So, again by perpendicular condition we get,

\overset{\to}{AP} \cdot \left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) = 0

$⇒ \lambda - 1 + \lambda + 10 + \lambda = 0$

$⇒ 3\lambda + 9 = 0$

$⇒ \lambda = - 3$

Hence,A\left(- 3, - 2, - 1\right) \text{and } P\left(1, - 9, 2\right)</p><p>So, by distance formula we get,AP = \sqrt{74}$</p></blockquote>$