Annuity & Perpetuity problem

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Suppose Joe has been paying $600$ from his monthly salary at the end of every month for the past $n$ years. After $n$ years of payments, he retires having purchased a perpetuity-due plan that begins right away, which gets him a continuous supply of annual salary payments at the start of each year. If $i>0$ is the underlying annual effective interest rate for all growth, and we know that $(1 + i)^n = 10.894481$, $a_{n, i}^{(12)} = 7.568285$, and $\ddot{a}_{\infty, i} = 8.884259$, what is the yearly pension Joe receives?

My attempt:

Let $X$ be the yearly pension payments.

First, notice the fact that $n$ is a yearly amount whereas Joe makes monthly payments. Moreover, these monthly payments are at the end of each month so we have an annuity-immediate.

Using the information we are given we can extract actual values:

$$\ddot{a}_{\infty, i} = \frac{1}{d} \implies d = \frac{1}{8.884259}$$

$$i = \frac{1}{1-d} - 1 = \frac{1}{1-\frac{1}{8.884259}} - 1 = 0.127$$

$$(1 + 0.127)^n = 10.894481 \implies n = 20 (years)$$

$$a_{n, i}^{(12)} = a_{n, i} \cdot \frac{i}{i^{(12)}} = \frac{1-v^n}{i} \cdot \frac{i}{i^{(12)}} = \frac{1-v^n}{i^{(12)}} \implies i^{(12)} = 0.12$$

, where $d$ is the discount rate and $i^{(12)}$ is the nominal yearly rate.

Now the future value of the annuity immediate with $240$ monthly payments is

$$F = 600 s_{240, \frac{i^{(12)}}{12}} = 600\frac{(1 + 0.12/12)^{240} - 1}{0.12/12} = 593624.68$$

The present value of the perpetuity due is

$$P = X\cdot \ddot{a}_{\infty, i} = X\cdot 8.884259$$

The future value and present value should be equal

$$8.884259X = 593624.68 \implies X = 66817.58$$

Is this correct? I did a lot of research but I am not sure if this is how to approach this problem. Any assistance is much appreciated.

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It is not necessary to use the $n$-year annuity-due to calculate $i^{(12)}$, since $$(1+i^{(12)}/12)^{12} = 1 + i,$$ and you already computed $i$. But you are correct; $i^{(12)} = 0.120009$ and $n = 20$. Then the remainder of your calculation is also correct.