Distance between two non-intersecting planes ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$ is $5$ units, where ${\mathrm{P}}_1$ is $2\mathrm{x}-3\mathrm{y}+6\mathrm{z}+26=0$ and ${\mathrm{P}}_2$ is $4\mathrm{x}+\mathrm{b}\mathrm{y}+\mathrm{c}\mathrm{z}+\mathrm{d}=0.$ The point $\mathrm{A}\left(-3,\mathrm{}0,-1\right)$ lies between the planes ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$, then the value of $3\mathrm{b}+4\mathrm{c}-5\mathrm{d}$ is equal to

• A

  $580$
• Correct answer
  B

  $120$
• C

  $-18$
• D

  $-120$

Hence, the correct answer is option: 'B'.