Let the plane P: 8x + α _ 1 y + α _ 2 z + 12 = 0 be parallel to the...

Given,

The plane $P: 8x + {\alpha}_{1}y + {\alpha}_{2}z + 12 = 0$ be parallel to the line $L: \dfrac{x + 2}{2} = \dfrac{y - 3}{3} = \dfrac{z + 4}{5} and the intercept of$P$on the$y$-axis is$1$,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/5f48fbac-6034-43c0-9084-6a2a53d555d3-9896384.png" style="width: 350px; height: 194px;" /></p><p> </p><p>Now normal of the plane will be perpendicular to the given line, so by perpendicular condition we get,</p><p>$⇒ 8 \times 2 + 3{\alpha}_{1} + 5{\alpha}_{2} = 0 \ldots \left(1\right)$</p><p>For$Y$intercept of plane put$x = z = 0$in equation of plane</p><p>$⇒ y = \dfrac{- 12}{{\alpha}_{1}} = 1 ⇒ {\alpha}_{1} = - 12$</p><p>Put${\alpha}_{1}$in equation \left(1\right)$ we get ${\alpha}_{2} = 4$

Hence, equation of plane $P: 2x - 3y + z + 3 = 0$

Now distance of point $A\left(- 2, 3, - 4\right)$ from the plane $P$ is $= \left|\dfrac{2\left(- 2\right) - 3\left(3\right) - 4 + 3}{\sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2} + {\left(1\right)}^{2}}}\right| = \sqrt{14}$.