If the equation of the plane containing the line x + 2y + 3z - 4 =...

Given,

The equation of the plane containing the line $x + 2y + 3z - 4 = 0 = 2x + y - z + 5$ and perpendicular to the plane\overset{\to}{r} = \left(\overset{^}{i} - \overset{^}{j}\right) + \lambda\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) + \mu ( \overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k} ) is $ax + by + cz = 4$,

Now let equation of required plane be,

$P: \left(x + 2y + 3z - 4\right) + \alpha\left(2x + y - z + 5\right) = 0$

$⇒ ( 2\alpha + 1 ) + \left(\alpha + 2\right)y + \left(3 - \alpha\right)z = 4 - 5\alpha$

So, the normal vector is given by,

\overset{\to}{{n}_{1}} = ( 2\alpha + 1 ) \overset{^}{i} + ( \alpha + 2 ) \overset{^}{j} + ( 3 - \alpha ) \overset{^}{k}

And normal of the plane \overset{\to}{r} = \left(\overset{^}{i} - \overset{^}{j}\right) + \lambda\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) + \mu ( \overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k} ) will be,

\overset{\to}{{n}_{2}} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 1 & 1 \\ 1 & - 2 & 3\end{cases}\right| = 5\overset{^}{j} - 2\overset{^}{j} - 3\overset{^}{k}

Now given they are perpendicular, so by perpendicular condition we get,

$\overset{\to}{{n}_{1}} \cdot \overset{\to}{{n}_{2}} = 0$

$⇒ 5\left(2\alpha + 1\right) - 2\left(\alpha + 2\right) - 3\left(3 - \alpha\right) = 0$

$⇒ 11\alpha + \left(- 8\right) = 0$

$⇒ \alpha = \dfrac{8}{11}$

Hence, equation of required plane will be,

$P: \dfrac{27}{11}x + \dfrac{30}{11}y + \dfrac{25}{11}z - \dfrac{4}{11} = 0$

$⇒ 27x + 30y + 25z = 4$

Now comparing with $ax + by + cz = 4$ we get,

$a = 27, b = 30, c = 25$

Hence, $a - b + c = 22$