Use line and plane distances to determine a parallel plane

$\ell:\mathbf r=(2,1,-2)+\lambda(0,-3,4)$. Substitution in π1 gives $2-2(1-3\lambda)-2(-2+4\lambda)=8$, or $4-2\lambda=8$. Hence λ=−2 and Q=(2,7,−10).

The normal is $(1,-2,-2)$ of magnitude 3 and line direction $(0,-3,4)$ has magnitude 5. Thus $\sin\theta=|(0,-3,4)\cdot(1,-2,-2)|/(5\cdot3)=2/15$. Also $PQ=(0,6,-8)$, $|PQ|=10$. With $\cos\theta=\sqrt{221}/15$, the requested projection length is $10\sqrt{221}/15=2\sqrt{221}/3$.

$PM=(0,4\alpha+3,3\alpha-4)$. Crossing with the unit direction gives $PM\times(0,-3,4)/5=(25\alpha,0,0)/5$, so the distance is $5|\alpha|$. Set this equal to 10, giving α=±2.

Put $OF=(2,1-3\lambda,-2+4\lambda)$ on ℓ, giving $MF=(0,-3\lambda-4\alpha-3,4\lambda-3\alpha+4)$. Dot with $(0,-3,4)$: $25\lambda+25=0$, hence λ=−1. Then $MF=(0,-4\alpha,-3\alpha)$, whose magnitude is $5|\alpha|=10$, giving α=±2 again.

For α>0 choose α=2, so M=(2,12,0). The plane through M parallel to π1 has normal $(1,-2,-2)$ and constant $-22$. Write π2:$\mathbf r\cdot(1,-2,-2)=\beta$. Its separation from the auxiliary plane must be 10, so $|\beta+22|/3=10$, giving β=8 or −52. The intended distinct plane π2 excludes β=8, which is π1 itself, leaving $x-2y-2z=-52$.

The distance from M=(2,12,0) to π2 is $|-22-\beta|/3=10$, again yielding β=8 or −52. Under the source’s distinct-parallel-plane interpretation, reject π1 and obtain $x-2y-2z=-52$. If coincident planes were allowed, the printed prompt would also permit π2=π1.